dung qua

a/ đk: x>0; \(x\ne1\)

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

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

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

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

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

vậy A =\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) với x>0; \(x\ne1\)

b/để A<0 thì: \(\frac{\sqrt{x}+1}{\sqrt{x}-1}< 0\)

mà x>0; \(x\ne1\)=> \(\sqrt{x}+1>0\)

nên để A<0 thì \(\sqrt{x}-1< 0\)=> x<1 kết hợp với đkxđ

=> \(0< x< 1\)

vậy \(0< x< 1\) thì A<0

Lúc đầu cành dưới có 14 con chim

 SO BE LA 999

SO LON LA 1010

cậu làm sai rồi

277 + 23

=300

K NHA

so lon la 75

so be la 25

ban nhe

dai qua viet ra kho lam

tk nhe@@@@@@@@@@^-^

đề bài sai rồi

thanks minh co viet sai

0^2+0^5=0