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Lời giải:
ĐK: $x\geq 0; x\neq 1$
\(A=\left[\frac{(\sqrt{x}-1)(x+2\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}+2}{(\sqrt{x}+1)(\sqrt{x}+2)}\right].\frac{\sqrt{x}-1}{(\sqrt{x}-1)(2\sqrt{x}+3)}\)
\(=\left(\frac{x+2\sqrt{x}+2}{\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{1}{2\sqrt{x}+3}=\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}.\frac{1}{2\sqrt{x}+3}=\frac{(\sqrt{x}+1)^2}{(\sqrt{x}+1)(2\sqrt{x}+3)}=\frac{\sqrt{x}+1}{2\sqrt{x}+3}\)
\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)
\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)
\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)
\(=>A=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left[\dfrac{\sqrt{x}+1-2}{x-1}\right]\)
\(=>A=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}\)
b,\(=>\dfrac{1}{\sqrt{x}}=\dfrac{1}{2}=>\sqrt{x}=2=>x=\sqrt{2}\left(tm\right)\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\text{x > 0, x ≠ 1}\)
\(A=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{x-1-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
a) \(P=\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}=\dfrac{\sqrt{5}+2+\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\dfrac{2\sqrt{5}}{\left(\sqrt{5}\right)^2-2^2}=2\sqrt{5}\)
b)\(Q=\left(1+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\cdot\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}-1+\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}}\)
\(Q=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{2}{\sqrt{x}-1}\)
Tick hộ nha
\(a,\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(l\right)\\x=-2\left(l\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\Leftrightarrow A\in\varnothing\\ b,\text{ý bạn là rút gọn A hả?}\\ A=\dfrac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x+4}{\left(x-2\right)\left(x+2\right)}\)
`a)(sqrtx-3)/(sqrtx-2)-(2sqrtx-1)/(sqrtx-1)+(x-2)/(x-3sqrtx+2)`
`=(x-4sqrtx+3-(2sqrtx-1)(sqrtx-2)+x-2)/(x-3sqrtx+2)`
`=(2x-4sqrtx+1-2x+5sqrtx-2)/(x-3sqrtx+2)`
`=(sqrtx-1)/(x-3sqrtx+2)`
`=1/(sqrtx-2)`
`b)((x+2)/(xsqrtx-1)-sqrtx/(x+sqrtx+1)+1/(1-sqrtx)):(sqrtx-1)/2`
`=((x+2-x+sqrtx-x-sqrtx-1)/(xsqrtx-1))*2/(sqrtx-1)`
`=(1-x)/(xsqrtx-1)*2/(sqrtx-1)`
`=(-2(sqrtx+1))/(x+sqrtx+1)`
a) Ta có: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-4\sqrt{x}+3-2x+4\sqrt{x}+\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
b) Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{-\sqrt{x}+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{-2\sqrt{x}-2}{x\sqrt{x}-1}\)
\(A=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x^2-2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x+2x^2+4x-2x^2+2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2}{x-2}\)