rút gọn các biểu thức sau:
A=$\frac{x^2+x}{x^2-2x+1}$:( $\frac{x+1}{x}$- $\frac{1}{1-x}$+ $\frac{2-x^2}{x^2-x}$ )
B=($\frac{2+x}{2-x}$- $\frac{4x^2}{x^2-4}$ -$\frac{2-x}{2+x}$ ):$\frac{x^2-3x}{2x^2-x^3}$
C=( $\frac{2x}{2x^2-5x+3}$ -$\frac{5}{2x-3}$):(3+ $\frac{2}{1-x}$ )
D=($\frac{x+1}{x-1}$- $\frac{x-1}{x+1}$ ):$\frac{2x}{5x-5}$
rút gọn các biểu thức sau: A=$\frac{x^2+x}{x^2-2x+1}$:( $\frac{x+1}{x}$- $\frac{1}{1-x}$+ $\frac{2-x^2}{x^2-x}$ ) B=($\frac{2+x}{2-x}$- $\frac{4x^2}{x
By Clara
Đáp án:
Giải thích các bước giải:
Đáp án:
$\begin{array}{l}
A = \dfrac{{{x^2} + x}}{{{x^2} – 2x + 1}}:\left( {\dfrac{{x + 1}}{x} – \dfrac{1}{{1 – x}} + \dfrac{{2 – {x^2}}}{{{x^2} – x}}} \right)\\
= \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x – 1} \right)}^2}}}:\left( {\dfrac{{\left( {x + 1} \right)\left( {x – 1} \right) + x + 2 – {x^2}}}{{x\left( {x – 1} \right)}}} \right)\\
= \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x – 1} \right)}^2}}}.\dfrac{{x\left( {x – 1} \right)}}{{{x^2} – 1 + x + 2 – {x^2}}}\\
= \dfrac{{x\left( {x + 1} \right)}}{{x – 1}}.\dfrac{x}{{1 + x}}\\
= \dfrac{{{x^2}}}{{x – 1}}\\
B = \left( {\dfrac{{2 + x}}{{2 – x}} – \dfrac{{4{x^2}}}{{{x^2} – 4}} – \dfrac{{2 – x}}{{2 + x}}} \right):\dfrac{{{x^2} – 3x}}{{2{x^2} – {x^3}}}\\
= \dfrac{{\left( {2 + x} \right)\left( {2 + x} \right) + 4{x^2} – \left( {2 – x} \right)\left( {2 – x} \right)}}{{\left( {2 – x} \right)\left( {2 + x} \right)}}.\dfrac{{{x^2}\left( {2 – x} \right)}}{{x\left( {x – 3} \right)}}\\
= \dfrac{{{x^2} + 4x + 4 + 4{x^2} – {x^2} + 4x – 4}}{{2 + x}}.\dfrac{x}{{x – 3}}\\
= \dfrac{{4{x^2} + 8x}}{{2 + x}}.\dfrac{x}{{x – 3}}\\
= \dfrac{{4{x^2}}}{{x – 3}}\\
C = \left( {\dfrac{{2x}}{{2{x^2} – 5x + 3}} – \dfrac{5}{{2x – 3}}} \right):\left( {3 + \dfrac{2}{{1 – x}}} \right)\\
= \dfrac{{2x – 5\left( {x – 1} \right)}}{{\left( {2x – 3} \right)\left( {x – 1} \right)}}:\dfrac{{3\left( {x – 1} \right) – 2}}{{x – 1}}\\
= \dfrac{{2x – 5x + 5}}{{\left( {2x – 3} \right)\left( {x – 1} \right)}}.\dfrac{{x – 1}}{{3x – 3 – 2}}\\
= \dfrac{{ – 3x + 5}}{{2x – 3}}.\dfrac{1}{{3x – 5}}\\
= \dfrac{1}{{3 – 2x}}\\
D = \left( {\dfrac{{x + 1}}{{x – 1}} – \dfrac{{x – 1}}{{x + 1}}} \right):\dfrac{{2x}}{{5x – 5}}\\
= \dfrac{{{{\left( {x + 1} \right)}^2} – {{\left( {x – 1} \right)}^2}}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{5\left( {x – 1} \right)}}{{2x}}\\
= \dfrac{{{x^2} + 2x + 1 – {x^2} + 2x – 1}}{{x + 1}}.\dfrac{5}{{2x}}\\
= \dfrac{{4x}}{{x + 1}}.\dfrac{5}{{2x}}\\
= \dfrac{{10}}{{x + 1}}
\end{array}$