điều kiện : \(\left\{{}\begin{matrix}10-x\ge0\\x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le10\\x\ge-3\end{matrix}\right.\Rightarrow-3\le x\le10\)
\(\Rightarrow\left(10-x\right)\left(x+3\right)\ge0\Leftrightarrow10x+30-x^2-3x\ge0\)
\(\Leftrightarrow-x^2+7x+30\ge0\) (tích 2 số dương)
ta có : \(\sqrt{10-x}+\sqrt{x+3}=5\Leftrightarrow\left(\sqrt{10-x}+\sqrt{x+3}\right)^2=25\)
\(\Leftrightarrow\left(\sqrt{10-x}\right)^2+2\sqrt{\left(10-x\right)\left(x+3\right)}+\left(\sqrt{x+3}\right)^2=25\)
\(\Leftrightarrow\left|10-x\right|+2\sqrt{10x+30-x^2-3x}+\left|x+3\right|=25\)
\(\Leftrightarrow10-x+x+3+2\sqrt{-x^2+7x+30}=25\) (vì \(\sqrt{10-x}và\sqrt{x+3}\ge0\) )
\(\Leftrightarrow13+2\sqrt{-x^2+7x+30}=25\Leftrightarrow2\sqrt{-x^2+7x+30}=12\)
\(\Leftrightarrow\left(2\sqrt{-x^2+7x+30}\right)^2=144\Leftrightarrow4\left|-x^2+7x+30\right|=144\)
\(\Leftrightarrow4\left(-x^2+7x+30\right)=144\Leftrightarrow-4x^2+28x+120=144\)
\(\Leftrightarrow144+4x^2-28x-120=0\Leftrightarrow4x^2-28x+24=0\)
ta có : \(a+b+c=4-28+24=0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=1\) ; \(x_2=\dfrac{24}{4}=6\)
vậy \(x=1;x=6\)
