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1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{4-x}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-1\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\left(\dfrac{2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}}\right)\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}}\)
\(=\dfrac{-4}{\sqrt{x}+2}\)
Lời giải:
a)
\(A=\left[\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\right].\frac{2-\sqrt{x}}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+2+2\sqrt{x}+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{2-\sqrt{x}}{\sqrt{x}}=\frac{4\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{2-\sqrt{x}}{\sqrt{x}}=\frac{-4}{\sqrt{x}+2}\)
b)
$A< -1\Leftrightarrow \frac{-4}{\sqrt{x}+2}+1< 0$
$\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+2}< 0$
$\Leftrightarrow \sqrt{x}-2< 0\Leftrightarrow 0\leq x< 4$
Kết hợp với ĐKXĐ suy ra $0< x< 4$
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-1\right)^2\left(\sqrt{x}-2\right)}\)
a) Ta có: \(Q=\left(\dfrac{x-1}{\sqrt{x}-1}-\dfrac{x\sqrt{x}-1}{x-1}\right):\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}-\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x-2\sqrt{x}+1+\sqrt{x}}{\sqrt{x}+1}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)^2\)
\(=\dfrac{x+2\sqrt{x}+1-x-\sqrt{x}-1}{\sqrt{x}+1}:\dfrac{\left(x-\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\left(x-\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}}{\left(x-\sqrt{x}+1\right)^2}\)
\(a,A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\left(dk:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x-1}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(b,x-3\sqrt{x}+2=0\Leftrightarrow x-\sqrt{x}-2\sqrt{x}+2=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Thay \(x=4\) vào A :
\(A=\dfrac{\sqrt{4}-1}{\sqrt{4}+1}=\dfrac{2-1}{2+1}=\dfrac{1}{3}\)
`P=(sqrtx/(sqrtx-1)+sqrtx/(x-1)):(2/x-(2-x)/(xsqrtx+x))`
`đk:x>0,x ne 1`
`P=((x+sqrtx+sqrtx)/(x-1)):(2/x+(x-2)/(x(sqrtx+1)))`
`=(x+2sqrtx)/(x-1):((2sqrtx+2+x-2)/(x(sqrtx+1)))`
`=(x+2sqrtx)/(x-1):(x+2sqrtx)/(x(sqrtx+1))`
`=(x+2sqrtx)/(x-1)*(x(sqrtx+1))/(x+2sqrtx)`
`=(x(sqrtx+1))/((sqrtx-1)(sqrtx+1))`
`=x/(sqrtx-1)`
`b)P>2`
`<=>x/(sqrtx-1)-2>0`
`<=>(x-2sqrtx+2)/(sqrtx-1)>0`
`<=>((sqrtx-1)^2+1)/(sqrtx-1)>0`
`<=>sqrtx-1>0`
`<=>x>1`
a) đk: x>0;x khác 1;0
P = \(\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{2}{x}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right]\)
= \(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x}{\sqrt{x}-1}\)
b) Để P > 2
<=> \(\dfrac{x}{\sqrt{x}-1}-2>0\)
<=> \(\dfrac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)
<=> \(\dfrac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)
<=> \(\sqrt{x}-1>0\)
<=> x > 1
a) Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
b) Ta có: \(x=4+2\sqrt{3}\)
\(\Leftrightarrow x=3+2\cdot\sqrt{3}\cdot1+1\)
hay \(x=\left(\sqrt{3}+1\right)^2\)
Thay \(x=\left(\sqrt{3}+1\right)^2\) vào biểu thức \(A=\dfrac{x-1}{\sqrt{x}}\), ta được:
\(A=\dfrac{\left(\sqrt{3}+1\right)^2-1}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\dfrac{4+2\sqrt{3}-1}{\sqrt{3}+1}\)
\(\Leftrightarrow A=\dfrac{\left(3+2\sqrt{3}\right)\left(\sqrt{3}-1\right)}{2}=\dfrac{3\sqrt{3}-3+6-2\sqrt{3}}{2}\)
\(\Leftrightarrow A=\dfrac{\sqrt{3}+3}{2}\)
Vậy: Khi \(x=4+2\sqrt{3}\) thì \(A=\dfrac{\sqrt{3}+3}{2}\)
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)
