Toán Lớp 9: Câu 1:
a) $\sqrt[]{4x+20}$ – 3$\sqrt[]{5}$ + x + 7$\sqrt[]{9x+45}$ = 20
b) $\left \{ {{5x-y=7} \atop {3x+y=9}} \right.$
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Toán Lớp 9: Câu 1:
a) $\sqrt[]{4x+20}$ – 3$\sqrt[]{5}$ + x + 7$\sqrt[]{9x+45}$ = 20
b) $\left \{ {{5x-y=7} \atop {3x+y=9}} \right.$
Comments ( 1 )
a)Dkxd:x \ge – 5\\
\sqrt {4x + 20} – 3\sqrt {5 + x} + 7\sqrt {9x + 45} = 20\\
\Leftrightarrow \sqrt {4\left( {x + 5} \right)} – 3\sqrt {x + 5} + 7.\sqrt {9\left( {x + 5} \right)} = 20\\
\Leftrightarrow 2\sqrt {x + 5} – 3\sqrt {x + 5} + 7.3\sqrt {x + 5} = 20\\
\Leftrightarrow 20\sqrt {x + 5} = 20\\
\Leftrightarrow \sqrt {x + 5} = 1\\
\Leftrightarrow x + 5 = 1\\
\Leftrightarrow x = – 4\left( {tm} \right)\\
Vậy\,x = – 4\\
b)\left\{ \begin{array}{l}
5x – y = 7\\
3x + y = 9
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
8x = 16\\
5x – y = 7
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 5x – 7
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 5.2 – 7 = 3
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {2;3} \right)
\end{array}$