Question

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

Answer 1

đk x>=1

\(< =>\dfrac{\sqrt{x^2-2x+5}^2-2^2}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0\)

\(< =>\dfrac{x^2-2x+1}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0\)

\(< =>\dfrac{\left(x-1\right)^2}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0\)

\(< =>\sqrt{x-1}\left(\dfrac{\sqrt{x-1}^3}{\sqrt{x^2-2x+5}+2}+1\right)=0\)

\(< =>x=1\left(tm\right)\)

nhận xét vì x>=1

\(\dfrac{\sqrt{x-1}^3}{\sqrt{x^2-2x+5}+2}+1>0\)

=> pt \(\dfrac{\sqrt{x-1}^3}{\sqrt{x^2-2x+5}+2}+1=0\) vô nghiệm

Answer 2

x = 4 * căn bậc hai ( 7 ) / 9 + 20 / 9