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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