Question

Giải PT:

\(2x^2-6x+10-5\left(x-2\right)\sqrt{x+1}=0\)

Answer 1

ĐKXĐ: \(x\ge-1\)

- Với \(x=-1\) ko phải nghiệm

- Với \(x>-1\)

\(\Leftrightarrow x^2-11x+24+\left(x-5\right)\left(x+7-5\sqrt{x+1}\right)=0\)

\(\Leftrightarrow x^2-11x+24+\frac{\left(x-5\right)\left(x^2-11x+24\right)}{x+7+5\sqrt{x+1}}=0\)

\(\Leftrightarrow\left(x^2-11x+24\right)\left(1+\frac{x-5}{x+7+5\sqrt{x+1}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-11x+24=0\Rightarrow x=...\\1+\frac{x-5}{x+7+5\sqrt{x+1}}=0\left(1\right)\end{matrix}\right.\)

Xét (1):

\(\Leftrightarrow x+7+5\sqrt{x+1}=5-x\)

\(\Leftrightarrow2\left(x+1\right)+5\sqrt{x+1}=0\) (vô nghiệm do \(x>-1\))

Vậy ...