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`a)đk:` $\begin{cases}\sqrt{x^2-2x} \ge 0\\x+\sqrt{x^2-2x} \ne 0\\x-\sqrt{x^2-2x} ne 0\\\end{cases}$
`<=>` $\begin{cases}x \ge 2\,or\,x<0\\x \ne 0\end{cases}$
`b)A=(x+sqrt{x^2-2x})/(x-sqrt{x^2-2x})-(x-sqrt{x^2-2x})/(x+sqrt{x^2+2x})`
`=((x+sqrt{x^2-2x})^2-(x-sqrt{x^2-2x})^2)/((x+sqrt{x^2-2x})(x-sqrt{x^2-2x}))`
`=(x^2+x^2-2x+2sqrt{x^2-2x}-x^2-x^2+2x+2sqrt{x^2-2x})/(x^2-x^2+2x)`
`=(4sqrt{x^2-2x})/(2x)`
`=(2sqrt{x^2-2x})/x`
`c)A<2`
`<=>2sqrt{x^2-2x}<2x`
`<=>sqrt{x^2-2x}<x(x>=2)`(BP 2 vế thì x>=2)
`<=>x^2-2x<x^2`
`<=>2x>0`
`<=>x>0`
`<=>x>=2`
Vậy `x>=2` thì `A<2`.
(Với x > 0; x 1; x4)
\(Đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\\left(\sqrt{x}-1\right)^2>0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x>1\end{cases}\Rightarrow}x>1}\)
\(C=\)\(\frac{1}{\sqrt{x}}+\frac{3}{x\sqrt{x}}+1+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}}+\frac{3}{x\sqrt{x}}+1+\frac{2}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{x\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{3\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{x\sqrt{x}\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{2x.\sqrt{x}}{x\sqrt{x}\left(\sqrt{x-1}\right)^2}\)
\(=x\left(\sqrt{x}-1\right)^2+3\left(\sqrt{x}-1\right)^2+x\sqrt{x}\left(\sqrt{x}-1\right)^2+2x.\sqrt{x}\)
.....
Ta có: \(P=\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}}{4-\sqrt{a}}\)
a) ĐKXĐ: \(a\ne4;a\ne16;a\ge0\)
\(P=\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}-\dfrac{4\sqrt{a}}{\sqrt{a}-4}\)
\(P=\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{4\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(P=\dfrac{a+3\sqrt{a}+2\sqrt{a}+6-a+2\sqrt{a}+\sqrt{a}-2-4\sqrt{a}}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\)
\(P=\dfrac{4\sqrt{a}+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\)
\(P=\dfrac{4\sqrt{a}+4}{a-4}\)
b) Thay x=9 vào P ta có:
\(P=\dfrac{4\cdot\sqrt{9}+4}{9-4}=\dfrac{16}{5}\)
c) \(P< 0\) khi:
\(\dfrac{4\sqrt{x}+4}{a-4}< 0\)
Mà: \(4\sqrt{x}+4>0\)
\(\Rightarrow a-4< 0\)
\(\Rightarrow a< 4\)
kết hợp với Đk ta có:
\(0\le x< 4\)
\(\sqrt{x^3}-1\)
\(\sqrt{x^3}-1\)