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\(a,\)\(T=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\sqrt{x}^3-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\)\(\frac{\sqrt{x}^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)\(-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
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
a) ĐKXĐ: x\(\ne\) 0;4
Ta có: Q= \(\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
= \(\frac{4\sqrt{x}\cdot\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\cdot\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)= \(\frac{4\sqrt{x}\cdot\left(2+\sqrt{x}\right)}{2+\sqrt{x}}\cdot\frac{-\sqrt{x}}{3-\sqrt{x}}\)=\(\frac{-4}{3-\sqrt{x}}\)=\(\frac{4}{\sqrt{x}-3}\)
b) Q=-1 => \(\frac{4}{\sqrt{x}-3}=-1\)
<=> \(4=3-\sqrt{x}\)
<=> \(\sqrt{x}=-1\) (vô lí)
Vậy ko tìm được x.
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)