Bài III.
a) Thay x=25 vào A ta có:
\(A=\dfrac{\sqrt{25}+2}{\sqrt{25}}=\dfrac{5+2}{5}=\dfrac{7}{5}\)
b) \(B=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(B=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{x+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
c) Ta có:
\(M=A\cdot B=\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)
\(\left|M\right|>M\) khi:
\(M< 0\)
\(\Rightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}< 0\)
\(\Rightarrow\sqrt{x}-1< 0\)
\(\Rightarrow\sqrt{x}< 1\)
\(\Rightarrow x< 1\)
\(\Rightarrow0< x< 1\)
1:
a: \(A=\sqrt{\left(3-\sqrt{5}\right)^2}+\dfrac{3}{4}\sqrt{80}-6\)
\(=3-\sqrt{5}+\dfrac{3}{4}\cdot4\sqrt{5}-6\)
\(=-3-\sqrt{5}+\sqrt{5}=-3\)
b: \(B=\dfrac{\sqrt{18}}{\sqrt{6}}+\dfrac{4}{\sqrt{5}-1}-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\)
\(=\sqrt{3}+\sqrt{5}+1-\sqrt{3}\)
\(=\sqrt{5}+1\)
c: \(C=sin^233^0-\dfrac{tan29}{cot61}-\dfrac{1}{2}\cdot cos^260^0+sin^257^0\)
\(=sin^233^0+cos^233^0-1-\dfrac{1}{2}\cdot\dfrac{1}{4}\)
\(=1-1-\dfrac{1}{8}=-\dfrac{1}{8}\)
