Toán Lớp 10: giải phương trình
$4\sqrt{x+4}-3\sqrt{x-3}=7$
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Toán Lớp 10: giải phương trình
$4\sqrt{x+4}-3\sqrt{x-3}=7$
Comments ( 1 )
4\sqrt {x + 4} – 3\sqrt {x – 3} = 7\\
\Leftrightarrow 4\sqrt {x + 4} = 7 + 3\sqrt {x – 3} \\
\Leftrightarrow 16{\sqrt {x + 4} ^2} = {\left( {7 + 3\sqrt {x – 3} } \right)^2}\\
\Leftrightarrow 16\left( {x + 4} \right) = 49 + 9\left( {x – 3} \right) + 7.3.2\sqrt {x – 3} \\
\Leftrightarrow 16x + 64 = 49 + 9x – 27 + 42\sqrt {x – 3} \\
\Leftrightarrow 42\sqrt {x – 3} = 7x + 42\\
\Leftrightarrow 6\sqrt {x – 3} = x + 6\\
\Leftrightarrow 36\left( {x – 3} \right) = {\left( {x + 6} \right)^2}\\
\Leftrightarrow 36x – 108 = {x^2} + 12x + 36\\
\Leftrightarrow {x^2} – 24x + 144 = 0\\
\Leftrightarrow {\left( {x – 12} \right)^2} = 0\\
\Leftrightarrow x = 12(tm)\\
\Rightarrow S = \left\{ {12} \right\}
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