ta có:\(\sqrt{8-x}+\sqrt{x+10}=x^2+2x+7\)(đk: \(-10\le x\le8\))
\(\Leftrightarrow\sqrt{8-x}+\sqrt{10+x}=\left(x+1\right)^2+6\)
áp dụng bđt bunhia cho bộ 4 số căn 8-x; căn 10+x; 1 ;1 ta có:
\(\left(8-x+10+x\right)\left(1+1\right)\ge\left(\sqrt{8-x}+\sqrt{10+x}\right)^2\)
\(\Leftrightarrow36\ge\left(\sqrt{8-x}+\sqrt{x+10}\right)^2\)
\(\Leftrightarrow\sqrt{x+10}+\sqrt{8-x}\le6\)
ta lại có: (x+1)2+6\(\ge6\)
mà \(\sqrt{8-x}+\sqrt{10+x}=\left(x+1\right)^2+6\)
=> \(\left(x+1\right)^2+6=6\Leftrightarrow x=-1\left(tm\right)\)
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