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\(\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
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H9
HT.Phong (9A5)
CTVHS
5 tháng 9 2023
\(E=\left(\dfrac{x\sqrt{x}}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) (ĐK: \(x\ne1;x>0\))
\(E=\left[\dfrac{\left(\sqrt{x}\right)^3-1^3}{x-\sqrt{x}}-\dfrac{\left(\sqrt{x}\right)^3+1^3}{x+\sqrt{x}}\right]+\left[\dfrac{x}{\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right]\left[\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(E=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]+\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(E=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right)+\dfrac{\left(\sqrt{x}\right)^2-1^2}{\sqrt{x}}\cdot\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(E=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\cdot\dfrac{2x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(E=\dfrac{2\sqrt{x}}{\sqrt{x}}+\dfrac{2x+2}{\sqrt{x}}\)
\(E=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
5 tháng 9 2023
https://hoc24.vn/cau-hoi/rut-gon-cac-bieu-thuc-sau21-a-leftdfracxsqrtx-1x-1-dfracx-1sqrtx-1right-leftsqrtx-dfracsqrtxsqrtx-1right-22-a-leftdfracxsqrtx.8357393071878
H9
HT.Phong (9A5)
CTVHS
16 tháng 8 2023
\(\left(\dfrac{1}{\sqrt{x}}-x\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)
\(=\dfrac{1-x\sqrt{x}}{\sqrt{x}}:\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)
\(=\dfrac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-\sqrt{x}}\)
\(=-\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=-\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
16 tháng 8 2023
\(\left(\dfrac{1}{\sqrt{x}}-x\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)=\dfrac{1-x\sqrt{x}}{\sqrt{x}}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+\left(1-\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{1-x\sqrt{x}}{\sqrt{x}}.\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-\sqrt{x}}=\dfrac{\left(1-x\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{1-x\sqrt{x}}{\sqrt{x}}\)
31 tháng 8 2021
Ta có: \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\left(1-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
31 tháng 8 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne1\\\sqrt{x}+1\ne0\left(luônđúng\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\left(1-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}-:\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+1-\sqrt{x}+1\right)\left(\sqrt{x}+1+\sqrt{x}-1\right)}{x-1}:\dfrac{2}{\sqrt{x}+1}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{2}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
5 tháng 5 2021
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: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
5 tháng 9 2023
21: ĐKXĐ: x>0; x<>1
\(A=\left(\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{x}\)
\(=\dfrac{-x+\sqrt{x}+2}{\sqrt{x}+1}\cdot\dfrac{1}{x}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}+2}{x}\)
22:
DKXĐ: x>0; x<>1
\(A=\dfrac{x-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1+\sqrt{x}+2-x}\)
\(=\dfrac{x}{\sqrt{x}+1}\)
23: ĐKXĐ: x>0; x<>4
\(A=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-4}\)
\(=\dfrac{-4\sqrt{x}+4}{4}=-\sqrt{x}+1\)
24: ĐKXĐ: x>=0; x<>1
\(A=\dfrac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+2}{x+\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
25:
ĐKXĐ: x>=0; x<>1
\(A=1:\dfrac{x+2\sqrt{x}-2-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{2x+\sqrt{x}-1-x+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x+\sqrt{x}}=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
27: ĐKXĐ: x>0; x<>4
\(P=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4x-8\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-1-2\sqrt{x}+1}\)
\(=\dfrac{4\left(x-2\sqrt{x}-2\right)}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\sqrt{x}}\)
\(=\dfrac{-4\left(x-2\sqrt{x}-2\right)}{\sqrt{x}+2}\)
31 tháng 8 2021
Ta có: \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
31 tháng 8 2021
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\dfrac{\sqrt{\left(x\right)^2}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
\(\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}}=\left(\dfrac{x-1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\)