Lời giải:
Đặt \(\sqrt{x-3}=t(t\geq 0)\)
PT tương đương với: \(\sqrt{8+t}+\sqrt{5-t}=5\)
\(\Leftrightarrow (\sqrt{8+t}-3)+(\sqrt{5-t}-2)=0\) (1)
\(\Leftrightarrow \frac{t-1}{\sqrt{8+t}+3}-\frac{t-1}{\sqrt{5-t}+2}=0\)
\(\Leftrightarrow (t-1)\left[\frac{1}{\sqrt{8+t}+3}-\frac{1}{\sqrt{5-t}+2}\right]=0\)
Vì \(t\geq 0\Rightarrow \sqrt{5-t}+2\leq \sqrt{5}+2< \sqrt{8}+3\leq \sqrt{8+t}+3\)
\(\Rightarrow \frac{1}{\sqrt{5-t}+2}>\frac{1}{\sqrt{8+t}+3}\Rightarrow \frac{1}{\sqrt{8+t}+3}-\frac{1}{\sqrt{5-t}+2}<0 \) (2)
Từ \((1);(2)\Rightarrow t-1=0\Leftrightarrow t=1\Leftrightarrow \sqrt{x-3}=1\Leftrightarrow x=4\)
(thỏa mãn)
\(\sqrt{8+\sqrt{x-3}}+\sqrt{5-\sqrt{x-3}}=5\)
\(\Leftrightarrow\left(\sqrt{x-3}\right)\left(\sqrt{8+1}+\sqrt{5-1}\right)=5\)
\(\Leftrightarrow\left(\sqrt{x-3}\right)\left(3+2\right)=5\)
\(\Leftrightarrow\left(\sqrt{x-3}\right)5=5\)
\(\Leftrightarrow\sqrt{x-3}=1\)
\(\Leftrightarrow x-3=1\)
\(\Rightarrow x=4\)
vậy x = 4
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