Question

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

a Tìm điều kiện của x để P xác định

b Rút gọn

Answer 1

đkxđ: \(\sqrt{x}\)-1 # 0

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

\(\sqrt{x}\)+1 #0 x# -1

x-1 # 0

rút gọn

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

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

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

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

chúc bạn học tốt

Answer 2

a) ĐKXĐ: x ≠ 1 ; x > 0

b) Ta có :

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

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

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

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

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

P= \(\frac{\sqrt{x}+1}{\sqrt{x}}.\frac{\sqrt{x}-1}{1}\)

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

Vậy P = \(\frac{x-1}{\sqrt{x}}\)