Question

\(x\ge0;x\ne1.\)Cho các biểu thức sau:

A= \(\frac{2\sqrt{x}+1}{3\sqrt{x}+1};\)

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

a, Tính giá trị của biểu thức A khi x=\(\frac{4}{\sqrt{3}-1}-\frac{4}{\sqrt{3}+1}\)

b, Rút gọn B c, tìm x để \(\frac{B}{A}>2\)

Answer 1

a, x = \(\frac{4\left(\sqrt{3}+1\right)}{3-1}-\frac{4\left(\sqrt{3}-1\right)}{3-1}\)

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

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

x = 4 (TMĐK)

=> A = \(\frac{2\sqrt{4}+1}{3\sqrt{4}+1}\)

=> A = \(\frac{5}{7}\)

Vậy x = \(\frac{4}{\sqrt{3}-1}-\frac{4}{\sqrt{3}+1}\) thì A = \(\frac{5}{7}\)

b, B = \(\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)

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

B = \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)

c, \(\frac{B}{A}>2\) <=> \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}:\frac{2\sqrt{x}+1}{3\sqrt{x}+1}\) > 2

<=> \(\frac{3\sqrt{x}+1}{\sqrt{x}+1}>2\)

<=> \(\frac{3\sqrt{x}+1}{\sqrt{x}+1}-2>0\)

<=> \(\frac{3\sqrt{x}+1-2\sqrt{x}-2}{\sqrt{x}+1}>0\)

<=> \(\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)

mà \(\sqrt{x}+1>0\) \(\forall\) \(x\in\) ĐKXĐ

=> \(\sqrt{x}-1>0\)

<=> \(\sqrt{x}>1\)

<=> \(x>1\)

Kết hợp ĐKXĐ : x \(\ge0\) ; x \(\ne\) 1

=> x > 1 thì \(\frac{B}{A}>2\)