Question

[truong hung dung]

 tim x biet \(\sqrt{x-1}\)+1=x

Answer 1

[Smile]

ĐKXĐ \(x\ge1\)

\(\Rightarrow\sqrt{x-1}=x-1\)

\(\Rightarrow x-1=x^2-2x+1\)

\(\Rightarrow x^2-3x+2=0\)

\(\Rightarrow x^2-x-2x+2=0\)

\(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Rightarrow x-1=0\Rightarrow x=1\) (nhận)

hoặc \(x-2=0\Rightarrow x=2\) (nhận)

Vậy x = 1 ; x = 2

Answer 2

[Lê Chí Cường]

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

=>\(\sqrt{x-1}=x-1\)

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

=>\(\sqrt{x-1}.\left(\sqrt{x-1}-1\right)=0\)

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

hoặc \(\sqrt{x-1}-1=0=>\sqrt{x-1}=1=>x-1=1=>x=2\)

Vậy x=1,2