1) Thay x=144 vào A ta có:
\(A=\dfrac{2\left(\sqrt{144}-1\right)}{\sqrt{144}+2}=\dfrac{2\cdot\left(12-1\right)}{12+2}=\dfrac{22}{14}=\dfrac{11}{7}\)
2) \(B=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(B=\left(\dfrac{15-\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{2}{\sqrt{x}+5}\right)\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(B=\left(\dfrac{15-\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right)\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(B=\left(\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right)\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(B=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(B=\dfrac{1}{\sqrt{x}-5}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(B=\dfrac{1}{\sqrt{x}+2}\)
3) \(B\ge A\) khi:
\(\dfrac{1}{\sqrt{x}+2}\ge\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}+2}\)
\(\Leftrightarrow1\ge2\sqrt{x}-2\)
\(\Leftrightarrow2\sqrt{x}\le1+2\)
\(\Leftrightarrow2\sqrt{x}\le3\)
\(\Leftrightarrow\sqrt{x}\le\dfrac{3}{2}\)
\(\Leftrightarrow x\le\dfrac{9}{4}\)
Kết hợp với đk ta kết luận
\(B\ge A\) khi \(0\le x\le\dfrac{9}{4}\)
1: Khi x=144 thì \(A=\dfrac{2\left(12-1\right)}{12+2}=\dfrac{2\cdot11}{14}=\dfrac{11}{7}\)
2: \(B=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{x-25}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(=\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}{x-25}\cdot\dfrac{1}{\sqrt{x}+2}=\dfrac{1}{\sqrt{x}+2}\)
3: B>=A
=>1>=2(căn x-1)
=>2căn x-2<=1
=>căn x<=3/2
=>0<=x<=9/4
