Giải phương trình :
ĐK : \(x\ge\frac{5}{2}\)
\(\sqrt{x-2}=2x-5\)
\(\Leftrightarrow x-2=\left(2x-5\right)^2\)
\(\Leftrightarrow4x^2-20x+25-x+2=0\)
\(\Leftrightarrow4x^2-21x+27=0\)
\(\Leftrightarrow4x^2-12x-9x+27=0\)
\(\Leftrightarrow4x\left(x-3\right)-9\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(4x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{9}{4}\end{matrix}\right.\)
Tìm giá trị nhỏ nhất :
a) \(A=\sqrt{x^2-4x+6}\)
\(A=\sqrt{x^2-4x+4+2}\)
\(A=\sqrt{\left(x-2\right)^2+2}\ge\sqrt{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=2\)
b) \(B=x-\sqrt{x-1}\)
\(B=x-1-2\sqrt{x-1}\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(B=\left(\sqrt{x-1}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x+1}=\frac{1}{2}\Leftrightarrow x+1=\frac{1}{4}\Leftrightarrow x=\frac{-3}{4}\)
