Bài 5 phương trình chứa ẩn ở mẫu

Gợi ý:

Điều kiện: \[x + 5 \ne 0 \Leftrightarrow x \ne – 5\]

\[\eqalign{
& \frac{{2x – 5}}{{x + 5}} = 3 \cr
& \Leftrightarrow \frac{{2x – 5}}{{x + 5}} = \frac{{3\left( {x + 5} \right)}}{{x + 5}} \cr
& \Leftrightarrow 2x – 5 = 3x + 15 \cr
& \Leftrightarrow x = – 20\,\left( {nhan} \right) \cr} \]

Gợi ý:

\[\eqalign{
& ĐK:x \ne 0 \cr
& \frac{{{x^2} – 6}}{x} = x + \frac{3}{2} \cr
& \Leftrightarrow \frac{{{x^2} – 6}}{x} = \frac{{2x + 3}}{2} \cr
& \Leftrightarrow \frac{{2\left( {{x^2} – 6} \right)}}{{2x}} = \frac{{x\left( {2x + 3} \right)}}{{2x}} \cr
& \Leftrightarrow 2{x^2} – 12 = 2{x^2} + 3x \cr
& \Leftrightarrow 3x = – 12 \Leftrightarrow x = – 4\,\,\left( {nhan} \right) \cr} \]

Gợi ý:

ĐK: \[x – 3 \ne 0 \Leftrightarrow x \ne 3\]

\[PT \Leftrightarrow \left( {{x^2} + 2x} \right) – \left( {3x + 6} \right) = 0\]

\[ \Leftrightarrow x\left( {x + 2} \right) – 3\left( {x + 2} \right) = 0\]

\[ \Leftrightarrow \left( {x + 2} \right)\left( {x – 3} \right) = 0\]

\[ \Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x – 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 2\\
x = 3\,\,\left( {loai} \right)
\end{array} \right.\]

Gợi ý:

\[{ĐK:3x + 2 \ne 0 \Leftrightarrow x \ne – \frac{2}{3}}\]

\[{\frac{5}{{3x + 2}} = 2x – 1}\]

\[{ \Leftrightarrow \frac{5}{{3x + 2}} = \frac{{\left( {2x – 1} \right)\left( {3x + 2} \right)}}{{3x + 2}}}\]

\[{ \Leftrightarrow 5 = \left( {2x – 1} \right)\left( {3x + 2} \right)}\]

\[{ \Leftrightarrow 5 = 6{x^2} + 4x – 3x – 2}\]

\[{ \Leftrightarrow 6{x^2} + x – 7 = 0}\]

\[{ \Leftrightarrow 6{x^2} – 6x + 7x – 7 = 0}\]

\[{ \Leftrightarrow \left( {6{x^2} – 6x} \right) + \left( {7x – 7} \right) = 0}\]

\[{ \Leftrightarrow 6x\left( {x – 1} \right) + 7\left( {x – 1} \right) = 0}\]

\[{ \Leftrightarrow \left( {x – 1} \right)\left( {6x + 7} \right) = 0}\]

\[{ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x – 1 = 0}\\
{6x + 7 = 0}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 1}\\
{x = – \frac{7}{6}}
\end{array}} \right.\left( {nhận{\mkern 1mu} hết} \right)}\]