Question

Tìm giá trị nhỏ nhất của mỗi biểu thức:

a,A=\(\frac{3}{2+\sqrt{2x-x^2+8}}\)

b,B=\(\sqrt{x-1-2\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}\)

Answer 1

a, \(2x-x^2+8=-\left(x^2-2x+1\right)+9=9-\left(x-1\right)^2\)

Vì \(-\left(x-1\right)^2\le0\Rightarrow9-\left(x-1\right)^2\le9\)

Suy ra \(\sqrt{2x-x^2+8}\le3\Rightarrow2+\sqrt{2x-x^2+8}\le5\)

\(\Rightarrow A=\frac{3}{2+\sqrt{2x-x^2+8}}\ge\frac{3}{5}\)

Dấu "=" xảy ra <=> x-1=0 <=> x=1

b, \(B=\sqrt{x-2-2\sqrt{x-2}+1}+\sqrt{x-2-2.3\sqrt{x-2}+9}\)

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

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

\(=\left|x-1\right|+\left|x+1\right|=\left|1-x\right|+\left|x+1\right|\ge\left|1-x+x+1\right|=2\)

Dấu "=" xảy ra khi \(\left(1-x\right)\left(x+1\right)\ge0\Leftrightarrow-1\le x\le1\)