Question

Gải phương trình :

\(x^2-3x+3-\sqrt{x-2}-\sqrt{7-x}=0\)

Answer 1

Lời giải:

ĐKXĐ: \(2\leq x\leq 7\)

PT \(\Leftrightarrow (x^2-3x)+(1-\sqrt{x-2})+(2-\sqrt{7-x})=0\)

\(\Leftrightarrow x(x-3)-\frac{x-3}{\sqrt{x-2}+1}+\frac{x-3}{\sqrt{7-x}+2}=0\)

\(\Leftrightarrow (x-3)\left[x-\frac{1}{\sqrt{x-2}+1}+\frac{1}{\sqrt{7-x}+2}\right]=0\)

Ta thấy: \(x\geq 2>1; \sqrt{x-2}+1\geq 1\Rightarrow \frac{1}{\sqrt{x-2}+1}\leq 1; \frac{1}{\sqrt{7-x}+2}>0\)

\(\Rightarrow x-\frac{1}{\sqrt{x-2}+1}+\frac{1}{\sqrt{7-x}+2}>0\)

\(\Rightarrow x-\frac{1}{\sqrt{x-2}+1}+\frac{1}{\sqrt{7-x}+2}\neq 0\)

Do đó: \(x-3=0\Leftrightarrow x=3\) (thỏa mãn)

Vậy PT có nghiệm $x=3$