Question

Câu 2, câu 3:

    

Answer 1

Câu 2:

1: \(P=\dfrac{x+\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{x-6\sqrt{x}+4}{x-4}\)

\(=\dfrac{x+\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x\sqrt{x}+3x+2\sqrt{x}-2x+4\sqrt{x}+\sqrt{x}-2+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

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

b: Thay \(x=9+4\sqrt{5}=\left(\sqrt{5}+2\right)^2\) vào P, ta được:

\(P=\dfrac{9+4\sqrt{5}+1}{\sqrt{\left(\sqrt{5}+2\right)^2}-2}\)

\(=\dfrac{10+4\sqrt{5}}{\sqrt{5}+2-2}=\dfrac{4\sqrt{5}+10}{\sqrt{5}}=4+2\sqrt{5}\)

Câu 3:

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

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

\(=\dfrac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-x-4\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-x-4\sqrt{x}-6-2x+12\sqrt{x}-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+8}{\sqrt{x}+1}\)

2: Thay \(x=4-2\sqrt{3}\) vào P, ta được:

\(P=\dfrac{4-2\sqrt{3}+8}{\sqrt{4-2\sqrt{3}}+1}\)

\(=\dfrac{12-2\sqrt{3}}{\sqrt{3}-1+1}=\dfrac{12-2\sqrt{3}}{\sqrt{3}}=4\sqrt{3}-2\)