Question

√x^2-2x+3=x-5

Answer 1

ĐKXĐ:

\(x-5>0\Leftrightarrow x>5\)

Ta có:

\(\sqrt{x^2-2x+3}=x-5\Leftrightarrow x^2-2x+3=\left(x-5\right)^2=x^2+25-10x\)

\(\Leftrightarrow x^2-2x+3-x^2-25+10x=0\)

\(\Leftrightarrow8x-22=0\Leftrightarrow x=\dfrac{22}{8}=\dfrac{11}{4}\left(ktmĐKXĐ\right)\)

Vậy phương trình vô nghiệm.

 

Answer 2

\(ĐK:x\ge5\\ PT\Leftrightarrow x^2-2x+3=x^2-10x+25\\ \Leftrightarrow8x=22\Leftrightarrow x=\dfrac{11}{4}\left(ktm\right)\\ \Leftrightarrow x\in\varnothing\)