Question

\(\sqrt{2x^2-4x+3}\)+\(\sqrt{3x^2-6x+7}\)=\(2-x^2+2x\)

Answer 1

ĐKXĐ: tự tìm

\(\sqrt{2x^2-4x+3}=\sqrt{2x^2-4x+2+1}=\sqrt{2.\left(x^2-2x+1\right)+1}\)

\(=\sqrt{2\left(x-1\right)^2+1}\ge1\)

\(\sqrt{3x^2-6x+7}=\sqrt{3x^2-6x+3+4}=\sqrt{3.\left(x^2-2x+1\right)+4}\)

\(=\sqrt{3.\left(x-1\right)^2+4}\ge2\)

\(\Rightarrow VT\ge1+2=3\text{,Dấu "=" xảy ra khi x=1}\)

\(2-x^2+2x=-x^2+2x-1+3=-\left(x^2-2x+1\right)+3\)

\(=-\left(x-1\right)^2+3\ge3\text{,Dấu "=" xảy ra khi x=1}\)

\(\text{Suy ra: }x=1\text{ là nghiệm của PT}\)