Bài tập 3 trang 71 SGK Toán 10 NC

a) ĐKXĐ: \(x \ne 1\)

Ta có 

\(\begin{array}{*{20}{l}}
{\begin{array}{*{20}{l}}
{x + \frac{1}{{x - 1}} = \frac{{2x - 1}}{{x - 1}}}\\
{ \Rightarrow x\left( {x - 1} \right) + 1 = 2x - 1}
\end{array}}\\
\begin{array}{l}
 \Leftrightarrow {x^2} - 3x + 2 = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 1\,\left( l \right)}\\
{x = 2\,\left( n \right)}
\end{array}} \right.
\end{array}
\end{array}\)

Vậy S = {2}

b) ĐKXĐ: \(x \ne 2\)

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

Vậy \(S = \emptyset \)

c) ĐKXĐ: \(x \ge 3\)

Ta có 

\(\begin{array}{*{20}{l}}
{\left( {{x^2} - 3x + 2} \right)\sqrt {x - 3}  = 0}\\
\begin{array}{l}
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\sqrt {x - 3}  = 0}\\
{{x^2} - 3x + 2 = 0}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 3{\mkern 1mu} {\mkern 1mu} \left( n \right)}\\
{x = 1{\mkern 1mu} {\mkern 1mu} \left( l \right)}\\
{x = 2{\mkern 1mu} {\mkern 1mu} \left( l \right)}
\end{array}} \right.
\end{array}
\end{array}\)

Vậy S = {3}

d) ĐKXĐ: \(x \ge -1\)

Ta có 

\(\begin{array}{l}
\left( {{x^2} - x - 2} \right)\sqrt {x + 1}  = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\sqrt {x + 1}  = 0}\\
{{x^2} - x - 2 = 0}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x =  - 1}\\
{x = 2}
\end{array}} \right.{\mkern 1mu} {\mkern 1mu} \left( n \right)
\end{array}\)

Vậy S = {- 1;2}

-- Mod Toán 10 HỌC247