Toán Lớp 10: Giải pt
$\frac{|3x-1|}{x+2}$ = |x-3|
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Toán Lớp 10: Giải pt
$\frac{|3x-1|}{x+2}$ = |x-3|
Comments ( 1 )
Dkxd:x \ne – 2\\
+ Khi:x \ge 3\\
\Leftrightarrow \dfrac{{3x – 1}}{{x + 2}} = x – 3\\
\Leftrightarrow 3x – 1 = \left( {x – 3} \right)\left( {x + 2} \right)\\
\Leftrightarrow 3x – 1 = {x^2} – x – 6\\
\Leftrightarrow {x^2} – 4x – 5 = 0\\
\Leftrightarrow \left( {x – 5} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow x = 5\left( {do:x \ge 3} \right)\\
+ Khi:\dfrac{1}{3} \le x < 3\\
\Leftrightarrow \dfrac{{3x – 1}}{{x + 2}} = 3 – x\\
\Leftrightarrow 3x – 1 = – \left( {x – 3} \right)\left( {x + 2} \right)\\
\Leftrightarrow 3x – 1 = – {x^2} + x + 6\\
\Leftrightarrow {x^2} + 2x – 7 = 0\\
\Leftrightarrow {\left( {x + 1} \right)^2} = 8\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\sqrt 2 – 1\left( {tm} \right)\\
x = – 2\sqrt 2 – 1\left( {ktm} \right)
\end{array} \right.\\
\Leftrightarrow x = 2\sqrt 2 – 1\\
+ Khi:x < \dfrac{1}{3};x \ne – 2\\
\Leftrightarrow \dfrac{{1 – 3x}}{{x + 2}} = 3 – x\\
\Leftrightarrow 3x – 1 = \left( {x – 3} \right)\left( {x + 2} \right)\\
\Leftrightarrow \left( {x – 5} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow x = – 1\\
Vậy\,x \in \left\{ { – 1;2\sqrt 2 – 1;5} \right\}
\end{array}$