A = ($\frac{1}{1-x}$ + $\frac{2}{x+1}$ – $\frac{5-x}{1-x^{2}}$) : $\frac{1-2x}{x^{2} – 1}$
Rút gọn biểu thức A
Tìm x để A > 0
B = ( $\frac{1}{3}$ + $\frac{3}{x^{2} – 3x}$) : ( $\frac{x^{2}}{27-3x^{2}}$ +$\frac{1}{x + 3}$)
Rút gọn biểu thức B
Tìm x để B < -1
A = ($\frac{1}{1-x}$ + $\frac{2}{x+1}$ – $\frac{5-x}{1-x^{2}}$) : $\frac{1-2x}{x^{2} – 1}$ Rút gọn biểu thức A Tìm x để A > 0 B = ( $\frac{1}{3}$ +
By Isabelle
Giải thích các bước giải:
1) ĐK: $x \ne \left\{ { \pm 1;\dfrac{1}{2}} \right\}$
a) Ta có:
$\begin{array}{l}
A = \left( {\dfrac{1}{{1 – x}} + \dfrac{2}{{x + 1}} – \dfrac{{5 – x}}{{1 – {x^2}}}} \right):\dfrac{{1 – 2x}}{{{x^2} – 1}}\\
= \left( {\dfrac{1}{{1 – x}} + \dfrac{2}{{1 + x}} + \dfrac{{x – 5}}{{1 – {x^2}}}} \right).\dfrac{{{x^2} – 1}}{{1 – 2x}}\\
= \dfrac{{1 + x + 2\left( {1 – x} \right) + x – 5}}{{1 – {x^2}}}.\dfrac{{{x^2} – 1}}{{1 – 2x}}\\
= \dfrac{{ – 2}}{{1 – {x^2}}}.\dfrac{{{x^2} – 1}}{{1 – 2x}}\\
= \dfrac{2}{{1 – 2x}}
\end{array}$
b) Để
$\begin{array}{l}
A > 0\\
\Leftrightarrow \dfrac{2}{{1 – 2x}} > 0\\
\Leftrightarrow 1 – 2x > 0\\
\Leftrightarrow x < \dfrac{1}{2}
\end{array}$
2) ĐK: $x \ne \left\{ {0; \pm 3} \right\}$
a) Ta có:
$\begin{array}{l}
B = \left( {\dfrac{1}{3} + \dfrac{3}{{{x^2} – 3x}}} \right):\left( {\dfrac{{{x^2}}}{{27 – 3{x^2}}} + \dfrac{1}{{x + 3}}} \right)\\
= \dfrac{{{x^2} – 3x + 9}}{{3\left( {{x^2} – 3x} \right)}}:\left( {\dfrac{{{x^2}}}{{3\left( {3 – x} \right)\left( {3 + x} \right)}} + \dfrac{1}{{x + 3}}} \right)\\
= \dfrac{{{x^2} – 3x + 9}}{{3\left( {{x^2} – 3x} \right)}}:\dfrac{{{x^2} + 3\left( {3 – x} \right)}}{{3\left( {3 – x} \right)\left( {3 + x} \right)}}\\
= \dfrac{{{x^2} – 3x + 9}}{{3x\left( {x – 3} \right)}}.\dfrac{{3\left( {3 – x} \right)\left( {3 + x} \right)}}{{{x^2} – 3x + 9}}\\
= \dfrac{{ – x – 3}}{x}
\end{array}$
b) Để $B<-1$
$\begin{array}{l}
\Leftrightarrow \dfrac{{ – x – 3}}{x} < – 1\\
\Leftrightarrow – 1 – \dfrac{3}{x} < – 1\\
\Leftrightarrow \dfrac{3}{x} > 0\\
\Leftrightarrow x > 0
\end{array}$