Bài tập 22 trang 84 SGK Toán 10 NC

a) Điều kiện: \(x \ne  - \frac{1}{2}\)

Ta có 

\(\begin{array}{l}
\frac{{2\left( {{x^2} - 1} \right)}}{{2x + 1}} = 2 - \frac{{x + 2}}{{2x + 1}}\\
 \Rightarrow 2\left( {{x^2} - 1} \right) = 2\left( {2x + 1} \right) - \left( {x + 2} \right)\\
 \Leftrightarrow 2{x^2} - 2 = 4x + 2 - x - 2\\
 \Leftrightarrow 2{x^2} - 3x - 2 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\,\,\left( n \right)\\
x =  - \frac{1}{2}\,\,\left( l \right)
\end{array} \right.
\end{array}\)

Vậy S = {2}

b) Điều kiện: \(\left\{ {\begin{array}{*{20}{l}}
{x \ne 1}\\
{x \ne  - \frac{5}{3}}
\end{array}} \right.\)

Ta có 

\(\begin{array}{*{20}{l}}
{\frac{{2x - 5}}{{x - 1}} = \frac{{5x - 3}}{{3x + 5}}}\\
\begin{array}{l}
 \Rightarrow \left( {2x - 5} \right)\left( {3x + 5} \right) = \left( {5x - 3} \right)\\
\,\,\,\,\,\,\,\,\,\,.\left( {x - 1} \right)
\end{array}\\
\begin{array}{l}
 \Leftrightarrow 6{x^2} + 10x - 15x - 25\\
\,\,\,\,\,\,\,\, = 5{x^2} - 5x - 3x + 3
\end{array}\\
\begin{array}{l}
 \Leftrightarrow {x^2} + 3x - 28 = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 4{\mkern 1mu} {\mkern 1mu} \left( n \right)}\\
{x =  - 7{\mkern 1mu} {\mkern 1mu} \left( n \right)}
\end{array}} \right.
\end{array}
\end{array}\)

Vậy S = {- 7;4}

-- Mod Toán 10 HỌC247