Đặt:\(\sqrt{x+9}=v;\sqrt{x+6}=u\)
Ta có: \(v+5u=5+vu\Leftrightarrow-v\left(u-1\right)+5\left(u-1\right)\Leftrightarrow\left(5-v\right)\left(u-1\right)\)
\(\left\{{}\begin{matrix}5-v=0\Leftrightarrow5=\sqrt{x+9}\Leftrightarrow x=16\left(N\right)\\u-1=0\Leftrightarrow\sqrt{x+6}=1\Leftrightarrow x=-5\left(N\right)\end{matrix}\right.ĐKXĐ:x>=-6\)
\(S=\left\{16,-5\right\}\)
Câu trên mình quên -5>-6
Đặt: \(\sqrt{x+9}=v;\sqrt{x+6}=u\)
Ta có: \(v+5u=5+vu\)
\(\Leftrightarrow v+5u-5-uv=0\)
\(\Leftrightarrow-v\left(u-1\right)+5\left(u-1\right)\)
\(\Leftrightarrow\left(5-v\right)\left(u-1\right)\)
\(\left\{{}\begin{matrix}5-v=0\Leftrightarrow5=\sqrt{x+9}\Leftrightarrow x=16\left(N\right)\\u-1=0\Leftrightarrow\sqrt{x+6}=1\Leftrightarrow x=-5\left(L\right)\end{matrix}\right.\) ĐKXĐ:\(x>=-6\)
\(S=\left\{16\right\}\)
\(\sqrt{x+9}+5\sqrt{x+\text{6}}=5+\sqrt{\left(x+9\right)\left(x+\text{6}\right)}\Leftrightarrow\sqrt{x+9}+5=5+\sqrt{x+9}\Leftrightarrow\sqrt{x+9}-\sqrt{x+9}=0\Leftrightarrow x+9-x-9=0\Leftrightarrow0=0\)
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