Toán Lớp 9: $\sqrt[]{2x +5}$- $\sqrt[]{3-x}$ =2
Đề bài: giải PT
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Toán Lớp 9: $\sqrt[]{2x +5}$- $\sqrt[]{3-x}$ =2
Đề bài: giải PT
Comments ( 2 )
\sqrt {2x + 5} – \sqrt {3 – x} = 2\\
\Leftrightarrow \sqrt {2x + 5} = 2 + \sqrt {3 – x} \\
\Leftrightarrow 2x + 5 = 4 + \left( {3 – x} \right) + 4\sqrt {3 – x} \\
\Leftrightarrow 2x + 5 = 7 – x + 4\sqrt {3 – x} \\
\Leftrightarrow 3x – 2 = 4\sqrt {3 – x} \\
\Leftrightarrow {\left( {3x – 2} \right)^2} = 16\left( {3 – x} \right)\\
\Leftrightarrow 9{x^2} – 12x + 4 = 48 – 16x\\
\Leftrightarrow 9{x^2} + 4x – 44 = 0\\
\Leftrightarrow 9{x^2} – 18x + 22x – 44 = 0\\
\Leftrightarrow 9x\left( {x – 2} \right) + 22\left( {x – 2} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {9x + 22} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2(tm)\\
x = – \dfrac{{22}}{9}(L)
\end{array} \right.\\
\Rightarrow S = \left\{ 2 \right\}
\end{array}$