Question

giai phuong trinh \(x^2-5x+4=2\sqrt{2x-4}\)

Answer 1

\(ĐK:x\ge2\)

\(x^2-5x+4=2\sqrt{2x-4}\)

<=>\(x^2-5x+4=2\sqrt{2\left(x-2\right)}\)

<=>\(x^2-5x+4+x-2+2=\left(x-2\right)+2\sqrt{2\left(x-2\right)}+2\)

<=>\(x^2-4x+4=\left(\sqrt{x-2}+2\right)^2\)

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

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

<=>\(\left(x-\sqrt{x-2}-4\right)\left(x+\sqrt{x-2}\right)=0\)

Xét \(x-\sqrt{x-2}-4=0\)

<=>\(x^2-8x+16=x-2\)

<=>\(x^2-9x+18=0\)

=> x=6;3(nhận)

Xet1\(x+\sqrt{x-2}=0\)

Do x\(\ge2\)=> pt vô nghiệm

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