Question

1) Giải phương trình : \(\sqrt{x-1}+\sqrt{x-4}=x^2-22\)

Answer 1

ĐKXĐ: \(x\ge4\)

\(\sqrt{x-1}-2+\sqrt{x-4}-1=x^2-25\)

\(\Leftrightarrow\dfrac{x-5}{\sqrt{x-1}+2}+\dfrac{x-5}{\sqrt{x-4}+1}=\left(x-5\right)\left(x+5\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\Rightarrow x=5\\\dfrac{1}{\sqrt{x-1}+2}+\dfrac{1}{\sqrt{x-4}+1}=x+5\left(1\right)\end{matrix}\right.\)

Xét (1):

Ta có \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-1}+2}< \dfrac{1}{2}\\\dfrac{1}{\sqrt{x-4}+1}< 1\end{matrix}\right.\) \(\Rightarrow VT< \dfrac{3}{2}\)

Mà \(x\ge4\Rightarrow x+5\ge9\Rightarrow VP\ge9\)

\(\Rightarrow VP>VT\Rightarrow\) pt (1) vô nghiệm với mọi \(x\ge4\)

Vậy pt có nghiệm duy nhất \(x=5\)