ta có :

Cho mình hỏi câu a của bạn phân số đầu tiên bạn vứt mất x ở mẫu của mik đâu rồi

a, ĐKXĐ: \(x>0;x\ne1;x\ne4\)

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

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

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

\(=\frac{\sqrt{x}-2}{\sqrt{x}}\)

giúp mk với mk cần gấp

a) M = \(\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}-\frac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)}+\frac{x^2-1}{\sqrt{x}\left(x-1\right)}\)(x>0;x khác 1)

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

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

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

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

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

b) M = 9/2

<=> \(\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\frac{9}{2}\)

<=> \(2x+4\sqrt{x}+2=9\sqrt{x}\)

<=> \(2x-5\sqrt{x}+2=0\)

<=> \(2x-\sqrt{x}-4\sqrt{x}+2=0\)

<=> \(\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

<=> \(\orbr{\begin{cases}x=\frac{1}{4}\\x=4\end{cases}\left(tm\right)}\)

Vậy...

c) \(\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)= \(\frac{x+2\sqrt{x}+1}{\sqrt{x}}=2+\frac{x+1}{\sqrt{x}}\ge2+\frac{2\sqrt{x}}{\sqrt{x}}=4\)

Dấu "=" xảy ra <=> x = 1.

Vậy M >=4