1) A= \(\dfrac{3}{\sqrt{7}-2}-\sqrt{11-4\sqrt{7}}=\dfrac{3}{\sqrt{7}-2}-\sqrt{\left(\sqrt{7}-2\right)^2}=\dfrac{3}{\sqrt{7}-2}-\left(\sqrt{7}-2\right)=\dfrac{3-\left(11-4\sqrt{7}\right)}{\sqrt{7}-2}=\dfrac{4\sqrt{7}-8}{\sqrt{7}-2}=4\)
2) a) B = \(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{2}{\sqrt{x}+2}\)
b) B = \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{4}\Leftrightarrow8=3\sqrt{x}+6\Leftrightarrow\sqrt{\text{x}}=\dfrac{2}{3}\Leftrightarrow\text{x}=\dfrac{4}{9}\)
Bài 1:
Ta có: \(A=\dfrac{3}{\sqrt{7}-2}-\sqrt{11-4\sqrt{7}}\)
\(=\dfrac{3\left(\sqrt{7}+2\right)}{3}-\left(\sqrt{7}-2\right)\)
\(=\sqrt{7}+2-\sqrt{7}+2=4\)
Bài 2:
a) Ta có: \(B=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}+2}\)
b) Để \(B=\dfrac{3}{4}\) thì \(3\sqrt{x}+6=8\)
\(\Leftrightarrow3\sqrt{x}=2\)
\(\Leftrightarrow x=\dfrac{4}{9}\)(thỏa ĐK)
