Question

tìm x

x+1=6\(-\sqrt{x+1}\)

 

Answer 1

ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow6-\sqrt{x+1}-x-1=0\\ \Leftrightarrow5-x=\sqrt{x+1}\\ \Leftrightarrow25-10x+x^2=x+1\left(x\le5\right)\\ \Leftrightarrow x^2-11x+24=0\\ \Leftrightarrow\left(x^2-3x\right)-\left(8x-24\right)=0\\ \Leftrightarrow x\left(x-3\right)-8\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=8\left(ktm\right)\end{matrix}\right.\)

Vậy \(x=3\)