Question

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

a) rút gọn P

b) tính giá trị của P khi x =\(\frac{2}{2-\sqrt{3}}-2\sqrt{3}\)

c) khi \(\sqrt{P}\) có nghĩa, hãy tìm GTNN của \(\sqrt{P}\)

Answer 1

a) Ta có: \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)

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

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

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

\(=\frac{x}{\sqrt{x}-1}\)

b) ĐKXĐ: \(0< x\ne1\)

Ta có: \(x=\frac{2}{2-\sqrt{3}}-2\sqrt{3}\)

\(=\frac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-2\sqrt{3}\)

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

=4(nhận)

Thay x=4 vào biểu thức \(P=\frac{x}{\sqrt{x}-1}\), ta được:

\(P=\frac{4}{\sqrt{4}-1}=\frac{4}{2-1}=4\)

Vậy: khi \(x=\frac{2}{2-\sqrt{3}}-2\sqrt{3}\) thì P=4

Answer 2

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

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

\(=\frac{x-4}{\sqrt{x}\left(x-1\right)}\)

Answer 3

b)\(x=\frac{2-2\sqrt{3}\left(2-\sqrt{2}\right)}{2-\sqrt{3}}=\frac{8-4\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2-\sqrt{3}\right)^2}{2-\sqrt{3}}=2-\sqrt{3}\)