Question

Giải phương trình: \(\left(\sqrt{x-1}+1\right)^3+2\sqrt{x-1}=2-x\)

Answer 1

Đặt \(x-1=a\) (a \(\ge\) 0)

=> pt : \(\left(\sqrt{a}+1\right)^3+2\sqrt{a}=2-\left(a+1\right)\)

\(\Leftrightarrow a\sqrt{a}+3\sqrt{a}+3a+1+2\sqrt{a}=2-a-1\)

\(\Leftrightarrow\left(5+a\right)\sqrt{a}+3a+a+1+1-2=0\)

\(\Leftrightarrow\left(5+a\right)\sqrt{a}+4a=0\)

\(\Leftrightarrow\sqrt{a}\cdot\left[\left(5+a\right)+4\sqrt{a}\right]=0\)

\(\Rightarrow\sqrt{a}=0\) (vì \(\left(5+a\right)+4\sqrt{a}>0\) )

\(\rightarrow\sqrt{x-1}=0\)

\(\rightarrow x=1\)

Vậy pt có nghiệm x = 1