{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $5(m+3x)(x+1)-4(1+2x)=85$ c\u00f3 m\u1ed9t nghi\u1ec7m $x=2$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $m=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","explain":" <span class='basic_left'> Thay $x=2$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho, ta \u0111\u01b0\u1ee3c: <br\/> $5(m+3x)(x+1)-4(1+2x)=85$ <br\/> $\\Leftrightarrow 5(m+3.2)(2+1)-4(1+2.2)=85$ <br\/> $\\Leftrightarrow 15(m+6)-4.5=85$ <br\/> $\\Leftrightarrow 15m+70=85$ <br\/> $\\Leftrightarrow 15m=85-70$ <br\/> $\\Leftrightarrow 15m=15$ <br\/> $\\Leftrightarrow m=1$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$ <\/span> <\/span> "}]}],"id_ques":831},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["7"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $4(m-x)(x+7)+4(x^2-x)=96$ c\u00f3 m\u1ed9t nghi\u1ec7m $x=25$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $m=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","explain":" <span class='basic_left'> Thay $x=25$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho, ta \u0111\u01b0\u1ee3c: <br\/> $4(m-x)(x+7)+4(x^2-x)=96$ <br\/> $\\Leftrightarrow 4(m-25)(25+7)+4(25^2-25)=96$ <br\/> $\\Leftrightarrow 4.32(m-25)+4.600=96$ <br\/> $\\Leftrightarrow 128m-3200+2400=96$ <br\/> $\\Leftrightarrow 128m-800=96$ <br\/> $\\Leftrightarrow 128m=96+800$ <br\/> $\\Leftrightarrow 128m=896$ <br\/> $\\Leftrightarrow m=7$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $7$ <\/span> <\/span> "}]}],"id_ques":832},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-40"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\left( \\dfrac{x+1}{39}+1 \\right)+\\left( \\dfrac{x+2}{38}+1 \\right)$$=\\left( \\dfrac{x+3}{37}+1 \\right)+\\left( \\dfrac{x+4}{36}+1 \\right)$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","hint":"Quy \u0111\u1ed3ng, r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c trong c\u00e1c ngo\u1eb7c c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 xu\u1ea5t hi\u1ec7n t\u1eed chung","explain":" <span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & \\left( \\dfrac{x+1}{39}+1 \\right)+\\left( \\dfrac{x+2}{38}+1 \\right)=\\left( \\dfrac{x+3}{37}+1 \\right)+\\left( \\dfrac{x+4}{36}+1 \\right) \\\\ & \\Leftrightarrow \\dfrac{x+40}{39}+\\dfrac{x+40}{38}=\\dfrac{x+40}{37}+\\dfrac{x+40}{36} \\\\ & \\Leftrightarrow \\dfrac{x+40}{39}+\\dfrac{x+40}{38}-\\dfrac{x+40}{37}-\\dfrac{x+40}{36}=0 \\\\ & \\Leftrightarrow \\left( x+40 \\right)\\left( \\dfrac{1}{39}+\\dfrac{1}{38}-\\dfrac{1}{37}-\\dfrac{1}{36} \\right)=0 \\\\ & Do\\,\\,\\dfrac{1}{39}+\\dfrac{1}{38}-\\dfrac{1}{37}-\\dfrac{1}{36}\\,\\,\\ne 0 \\\\ & \\Rightarrow x+40=0 \\\\ & \\Leftrightarrow x=-40 \\\\ \\end{align}$ <br\/> Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x = -40$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-40$ <\/span> <\/span> "}]}],"id_ques":833},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["89"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x-12}{77}+\\dfrac{x-11}{78}$$=\\dfrac{x-74}{15}+\\dfrac{x-73}{16}$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","hint":"C\u1ed9ng m\u1ed7i ph\u00e2n th\u1ee9c trong ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean v\u1edbi $-1$ ta \u0111\u01b0\u1ee3c nh\u00e2n t\u1eed chung","explain":" <span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{x-12}{77}+\\dfrac{x-11}{78}=\\dfrac{x-74}{15}+\\dfrac{x-73}{16} \\\\ & \\Leftrightarrow \\left( \\dfrac{x-12}{77}-1 \\right)+\\left( \\dfrac{x-11}{78}-1 \\right)=\\left( \\dfrac{x-74}{15}-1 \\right)+\\left( \\dfrac{x-73}{16}-1 \\right) \\\\ & \\Leftrightarrow \\dfrac{x-89}{77}+\\dfrac{x-89}{78}=\\dfrac{x-89}{15}+\\dfrac{x-89}{16} \\\\ & \\Leftrightarrow \\dfrac{x-89}{77}+\\dfrac{x-89}{78}-\\dfrac{x-89}{15}-\\dfrac{x-89}{16}=0 \\\\ & \\Leftrightarrow \\left( x-89 \\right)\\left( \\dfrac{1}{77}+\\dfrac{1}{78}-\\dfrac{1}{15}-\\dfrac{1}{16} \\right)=0 \\\\ & Do\\,\\,\\dfrac{1}{77}+\\dfrac{1}{78}-\\dfrac{1}{15}-\\dfrac{1}{16}\\,\\,\\,\\ne 0 \\\\ & \\Leftrightarrow x-89=0 \\\\ & \\Leftrightarrow x=89 \\\\ \\end{align}$ <br\/> Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x = 89$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $89$ <\/span> <br\/> <b> L\u01b0u \u00fd: <\/b> <i>C\u1ed9ng c\u1ea3 hai v\u1ebf c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi c\u00f9ng m\u1ed9t s\u1ed1 ta \u0111\u01b0\u1ee3c m\u1ed9t ph\u01b0\u01a1ng tr\u00ecnh m\u1edbi t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho. <\/i> <\/span> "}]}],"id_ques":834},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" <span class='basic_left'> Ph\u01b0\u01a1ng tr\u00ecnh ${{\\left( x+2 \\right)}^{3}}+{{\\left( x-2 \\right)}^{3}}-{{x}^{3}}$$=x\\left( {{x}^{2}}-4 \\right)-2$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 <\/span>","select":["A. $x=-3$","B. $x=5$","C. $x=-\\dfrac{2}{7}$","D. $x=-\\dfrac{1}{14}$"],"explain":" <span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & {{\\left( x+2 \\right)}^{3}}+{{\\left( x-2 \\right)}^{3}}-{{x}^{3}}=x\\left( {{x}^{2}}-4 \\right)-2 \\\\ & \\Leftrightarrow {{x}^{3}}+6{{x}^{2}}+12x+8+{{x}^{3}}-6{{x}^{2}}+12x-8-{{x}^{3}}={{x}^{3}}-4x-2 \\\\ & \\Leftrightarrow {{x}^{3}}+24x={{x}^{3}}-4x-2 \\\\ & \\Leftrightarrow 24x+4x=-2 \\\\ & \\Leftrightarrow 28x=-2 \\\\ & \\Leftrightarrow x=-\\dfrac{2}{28} \\\\ & \\Leftrightarrow x=-\\dfrac{1}{14} \\\\ \\end{align}$ <br\/> Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $x=-\\dfrac{1}{14}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span><\/span> ","column":2}]}],"id_ques":835},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho hai bi\u1ec3u th\u1ee9c $A$ v\u00e0 $B$ sau: <br\/> $A=(x-3)(x+4)-2(3x-2);\\,\\,\\,B={{(x-4)}^{2}}$ <br\/> \u0110\u1ec3 $A=B$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0: <\/span>","select":["A. $8$","B. $4$","C. $-3$","D. $-8$"],"explain":" <span class='basic_left'> Cho $A=B$ ta \u0111\u01b0\u1ee3c: <br\/> $\\begin{align} & (x-3)(x+4)-2(3x-2)\\,\\,\\,={{(x-4)}^{2}} \\\\ & \\Leftrightarrow {{x}^{2}}+4x-3x-12-6x+4={{x}^{2}}-8x+16 \\\\ & \\Leftrightarrow -5x-8=-8x+16 \\\\ & \\Leftrightarrow -5x+8x=16+8 \\\\ & \\Leftrightarrow 3x=24 \\\\ & \\Leftrightarrow x=8 \\\\ \\end{align}$ <br\/> \u0110\u1ec3 $A=B$ th\u00ec $x=8$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span><\/span> ","column":4}]}],"id_ques":836},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho hai bi\u1ec3u th\u1ee9c $A$ v\u00e0 $B$ sau: <br\/> $A=(x-1)({{x}^{2}}+x+1)-2x;\\,\\,\\,B=x(x-1)(x+1)$ <br\/> \u0110\u1ec3 $A=B$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0: <\/span>","select":["A. $-3$","B. $-1$","C. $\\dfrac{1}{2}$","D. $\\dfrac{2}{3}$"],"explain":" <span class='basic_left'> Cho $A=B$ ta \u0111\u01b0\u1ee3c: <br\/> $\\begin{align} & (x-1)({{x}^{2}}+x+1)-2x=x(x-1)(x+1) \\\\ & \\Leftrightarrow {{x}^{3}}-1-2x=x({{x}^{2}}-1) \\\\ & \\Leftrightarrow -1-2x=-x \\\\ & \\Leftrightarrow -2x+x=1 \\\\ & \\Leftrightarrow -x=1 \\\\ &\\Leftrightarrow x=-1 \\\\ \\end{align}$ <br\/> \u0110\u1ec3 $A=B$ th\u00ec $x=-1$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span><\/span> ","column":2}]}],"id_ques":837},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho hai bi\u1ec3u th\u1ee9c $A$ v\u00e0 $B$ sau: <br\/> $A=\\dfrac{x+4}{5}+\\dfrac{3x+2}{10}$$\\,\\,;\\,\\,B=\\dfrac{x-1}{3}-7$ <br\/> \u0110\u1ec3 $A=B$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0: <\/span>","select":["A. $10$","B. $25$","C. $-25$","D. $-50$"],"explain":" <span class='basic_left'> Cho $A=B$ ta \u0111\u01b0\u1ee3c: <br\/> $\\begin{align} & \\dfrac{x+4}{5}+\\dfrac{3x+2}{10}=\\dfrac{x-1}{3}-7 \\\\ & \\Leftrightarrow \\left( x+4 \\right)6+\\left( 3x+2 \\right)3=\\left( x-1 \\right)10-7.30 \\\\ & \\Leftrightarrow 6x+24+9x+6=10x-10-210 \\\\ & \\Leftrightarrow 15x+30=10x-220 \\\\ & \\Leftrightarrow 15x-10x=-220-30 \\\\ & \\Leftrightarrow 5x=-250 \\\\ & \\Leftrightarrow x=-50 \\\\ \\end{align}$ <br\/> \u0110\u1ec3 $A=B$ th\u00ec $x=-50$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span><\/span> ","column":2}]}],"id_ques":838},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["10"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh <br\/> $\\left( \\dfrac{1}{1.101}+\\dfrac{1}{2.102}+\\dfrac{1}{3.103}+...+\\dfrac{1}{10.110} \\right)x$$=\\dfrac{1}{1.11}+\\dfrac{1}{2.12}+...+\\dfrac{1}{100.110}$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","hint":" \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c $\\dfrac{1}{n(n+a)}= \\dfrac{1}{a}\\left(\\dfrac{1}{n}-\\dfrac{1}{n+a}\\right)$ \u0111\u1ec3 r\u00fat g\u1ecdn v\u1ebf tr\u00e1i v\u00e0 v\u1ebf ph\u1ea3i","explain":" <span class='basic_left'> $\\left( \\dfrac{1}{1.101}+\\dfrac{1}{2.102}+\\dfrac{1}{3.103}+...+\\dfrac{1}{10.110} \\right)x=\\dfrac{1}{1.11}+\\dfrac{1}{2.12}+...+\\dfrac{1}{100.110}$ (1) <br\/> \u0110\u1eb7t $A=\\dfrac{1}{1.101}+\\dfrac{1}{2.102}+\\dfrac{1}{3.103}+...+\\dfrac{1}{10.110}$ <br\/> $\\begin{align} & =\\dfrac{1}{100}\\left( \\dfrac{1}{1}-\\dfrac{1}{101}+\\dfrac{1}{2}-\\dfrac{1}{102}+\\dfrac{1}{3}-\\dfrac{1}{103}+...+\\dfrac{1}{10}-\\dfrac{1}{110} \\right) \\\\ & =\\dfrac{1}{100}\\left[ \\left( 1+\\dfrac{1}{2}+\\dfrac{1}{3}+...+\\dfrac{1}{10} \\right)-\\left( \\dfrac{1}{101}+\\dfrac{1}{102}+\\dfrac{1}{103}+...+\\dfrac{1}{110} \\right) \\right] \\\\ \\end{align}$ <br\/> \u0110\u1eb7t $B=\\dfrac{1}{1.11}+\\dfrac{1}{2.12}+...+\\dfrac{1}{100.110}$ <br\/> $\\begin{align} & =\\dfrac{1}{10}\\left( \\dfrac{1}{1}-\\dfrac{1}{11}+\\dfrac{1}{2}-\\dfrac{1}{12}+...+\\dfrac{1}{100}-\\dfrac{1}{110} \\right) \\\\ & =\\dfrac{1}{10}\\left[ \\left( 1+\\dfrac{1}{2}+\\dfrac{1}{3}+...+\\dfrac{1}{100} \\right)-\\left( \\dfrac{1}{11}+\\dfrac{1}{12}+...+\\dfrac{1}{110} \\right) \\right] \\\\ & =\\dfrac{1}{10}\\left[ \\left( 1+\\dfrac{1}{2}+\\dfrac{1}{3}+...+\\dfrac{1}{10} \\right)-\\left( \\dfrac{1}{101}+\\dfrac{1}{102}+...+\\dfrac{1}{110} \\right) \\right] \\\\ \\end{align}$ <br\/> Do \u0111\u00f3: $10A=B$ <br\/> Ph\u01b0\u01a1ng tr\u00ecnh (1) $\\Leftrightarrow A.x=B\\Leftrightarrow A.x=10.A\\Leftrightarrow x=10$ <br\/> V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\left\\{ 10 \\right\\}$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $10$ <\/span> <br\/> <b> L\u01b0u \u00fd: <\/b> $\\dfrac{1}{n(n+a)}$$= \\dfrac{1}{a}.\\dfrac{a}{n(n+a)}$$= \\dfrac{1}{a}.\\dfrac{(n + a) - n}{n(n+a)}$$= \\dfrac{1}{a}.\\left(\\dfrac{n+a}{n(n+a)}-\\dfrac{n}{n(n+a)}\\right)$$= \\dfrac{1}{a}.\\left[\\dfrac{1}{n}-\\dfrac{1}{n+a}\\right]$ <br\/> V\u1eady ta c\u00f3 c\u00f4ng th\u1ee9c: $\\dfrac{1}{n(n+a)}= \\dfrac{1}{a}\\left(\\dfrac{1}{n}-\\dfrac{1}{n+a}\\right)$<\/span> "}]}],"id_ques":839},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" <span class='basic_left'> \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{2\\left( x-4 \\right)}{3}-\\dfrac{9x-2}{4}=\\dfrac{3x+1}{2}+\\dfrac{3x-1}{6}$ v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: <\/span>","select":["A. $43x+22=0$","B. $15x+30=0$","C. $43x+30=0$","D. $15x+22=0$"],"explain":" <span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{2\\left( x-4 \\right)}{3}-\\dfrac{9x-2}{4}=\\dfrac{3x+1}{2}+\\dfrac{3x-1}{6} \\\\ & \\Leftrightarrow 8\\left( x-4 \\right)-3\\left( 9x-2 \\right)=6\\left( 3x+1 \\right)+2\\left( 3x-1 \\right) \\\\ & \\Leftrightarrow 8x-32-27x+6=18x+6+6x-2 \\\\ & \\Leftrightarrow -19x-26=24x+4 \\\\ & \\Leftrightarrow -19x-24x=4+26 \\\\ & \\Leftrightarrow -43x=30 \\\\ & \\Leftrightarrow 43x+30=0 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C <\/span><\/span> ","column":4}]}],"id_ques":840}],"lesson":{"save":0,"level":3}}