Giải bài 6.42 trang 26 SGK Toán 8 tập 2 - Kết nối tri thức

Đề bài

Rút gọn biểu thức sau:

a) $\frac{2}{{3{{x}}}} + \frac{x}{{x - 1}} + \frac{{6{{{x}}^2} - 4}}{{2{{x}}\left( {1 - x} \right)}}$

b) $\frac{{{x^3} + 1}}{{1 - {x^3}}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}$

c) $\left( {\frac{2}{{x + 2}} - \frac{2}{{1 - x}}} \right).\frac{{{x^2} - 4}}{{4{{{x}}^2} - 1}}$

d) $1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)$

Lời giải chi tiết

a) $\frac{2}{3x}+\frac{x}{x-1}+\frac{6{{x}^{2}}-4}{2x\left( 1-x \right)}$

$=\frac{2}{3x}+\frac{x}{x-1}+\frac{3{{x}^{2}}-2}{x\left( 1-x \right)}$

$=\frac{2}{3x}+\frac{x}{x-1}-\frac{3{{x}^{2}}-2}{x\left( x-1 \right)}$

$=\frac{2\left( x-1 \right)+x.3x-3\left( 3{{x}^{2}}-2 \right)}{3x\left( x-1 \right)}$

$=\frac{2x-2+3{{x}^{2}}-9{{x}^{2}}+6}{3x\left( x-1 \right)}$

$=\frac{-6{{x}^{2}}+2x+4}{3x\left( x-1 \right)}$

$=\frac{-2\left( 3{{x}^{2}}-x-2 \right)}{3x\left( x-1 \right)}$

$=\frac{-2\left( 3{{x}^{2}}-3x+2x-2 \right)}{3x\left( x-1 \right)}$

$=\frac{-2\left[ 3x\left( x-1 \right)+2\left( x-1 \right) \right]}{3x\left( x-1 \right)}$

$=\frac{-2\left( 3x+2 \right)\left( x-1 \right)}{3x\left( x-1 \right)}$

$=\frac{-2\left( 3x+2 \right)}{3x}$

b) $\frac{{{x^3} + 1}}{{1 - {x^3}}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}$

$= \frac{{ - {x^3} - 1}}{{{x^3} - 1}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}$

$= \frac{{ - {x^3} - 1 + x\left( {{x^2} + x + 1} \right) - \left( {{x^2} - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$

$= \frac{{ - {x^3} - 1 + {x^3} + {x^2} + x - {x^2} + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$ $= \frac{x}{{{x^3} - 1}}$

c) Ta có: 

$\frac{2}{{x + 2}} - \frac{2}{{1 - x}}$

$= \frac{{2\left( {1 - x} \right) - 2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}$

$= \frac{{2 - 2{{x}} - 2{{x}} - 4}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}$

$= \frac{{ - 4x - 2}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}$

$= \frac{{2\left( {2x + 1} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}$;

$\frac{{{x^2} - 4}}{{4{{{x}}^2} - 1}}$ $= \frac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}}$.

Do đó

$\left( {\frac{2}{{x + 2}} - \frac{2}{{1 - x}}} \right).\frac{{{x^2} - 4}}{{4{{{x}}^2} - 1}}$

$= \frac{{2\left( {2x + 1} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}.\frac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}}$

$= \frac{{2(x - 2)}}{{\left( {2x - 1} \right)\left( {x - 1} \right)}}$

d) $1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)$

$= 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)$

$= 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}.\frac{{1 + x - 1}}{{1 - {x^2}}}$

$= 1 + \frac{{x\left( {{x^2} - 1} \right)}}{{{x^2} + 1}}.\frac{x}{{1 - {x^2}}}$

$= 1 + \frac{{ - {x^2}\left( {{x^2} - 1} \right)}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}$

$= 1 + \frac{{ - {x^2}}}{{{x^2} + 1}}$

$= \frac{{{x^2} + 1 - {x^2}}}{{{x^2} + 1}}$

$= \frac{1}{{{x^2} + 1}}$