Question

Answer 1

a: Thay x=100 vào B, ta được:

\(B=\dfrac{10+2}{10+1}=\dfrac{12}{11}\)

b: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{x-4}-\dfrac{1}{2-\sqrt{x}}\)

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

\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+1}{x-4}\)

\(=\dfrac{2x-5\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

c: \(M=A\cdot B=\dfrac{2x-5\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

\(=\dfrac{2x-5\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)