Rút gọn:
\(A=\left(\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\dfrac{\sqrt{a}-2}{a-1}\right).\dfrac{\sqrt{a}+1}{\sqrt{a}}\left(a>0,a\ne1\right)\)
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9 tháng 2 2022
b. \(=\left(\dfrac{\sqrt{a}-a+a\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right):\left(\dfrac{2\sqrt{a}}{1+\sqrt{a}}\right)\)
\(=\left(\dfrac{2\sqrt{a}}{1-\sqrt{a}}\right):\left(\dfrac{2\sqrt{a}}{1+\sqrt{a}}\right)\)
\(=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\)
\(=1-a\)
HP
9 tháng 2 2022
\(a.\sqrt{8}-2\sqrt{50}+\sqrt{18}=2\sqrt{2}-10\sqrt{2}+3\sqrt{2}=\sqrt{2}\left(2-10+3\right)=-5\sqrt{2}\)
\(b.\left(\dfrac{\sqrt{a}-a}{1-\sqrt{a}}+\sqrt{a}\right):\dfrac{2\sqrt{a}}{1+\sqrt{a}}\left(đk:a\ge0;a\ne1\right)\)
\(=\left(\sqrt{a}+\sqrt{a}\right).\dfrac{1+\sqrt{a}}{2\sqrt{a}}\)
\(=2\sqrt{a}.\dfrac{1+\sqrt{a}}{2\sqrt{a}}\)
\(=1+\sqrt{a}\)
(Chỗ điều kiện bài b mik thấy a = 0 cũng có thể là nghiệm nên mik sửa lại nhé)
HP
17 tháng 12 2020
\(P=\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{a-\sqrt{a}}\right):\dfrac{1}{\sqrt{a}-1}\)
\(=\left[\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\sqrt{a}}+\dfrac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}\right].\left(\sqrt{a}-1\right)\)
\(=\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\left(\sqrt{a}-1\right)=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
5 tháng 5 2021
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
5 tháng 5 2021
Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
26 tháng 7 2021
a) Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(1-\dfrac{1}{\sqrt{a}}\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{-\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}-1}{\sqrt{2}}\)
\(=\dfrac{-2}{\sqrt{a}+1}\)
b) Ta có: \(P=\dfrac{-1}{2}\)
nên \(\dfrac{2}{\sqrt{a}+1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{a}+1=4\)
\(\Leftrightarrow a=9\)(thỏa ĐK)
28 tháng 6 2021
e) Ta có: \(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\cdot\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(2x-3\sqrt{x}+1\right)-x\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{2x\sqrt{x}-3x+\sqrt{x}-x\sqrt{x}-x-\sqrt{x}}{x+\sqrt{x}+1}\cdot\dfrac{1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-4x}{x+\sqrt{x}+1}\cdot\dfrac{1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-4x+\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-4x+x\sqrt{x}+x+\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2x\sqrt{x}-3x+\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}\)
28 tháng 6 2021
m) Ta có: \(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2}{a-1}\right)\)
\(=\left(\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-1-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-3}\)
\(=\left(\sqrt{a}-1\right)\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
26 tháng 9 2021
a) \(M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
b) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\)
c) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=\dfrac{\sqrt{3-2\sqrt{2}}-1}{\sqrt{3-2\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-1}{\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}\)
26 tháng 9 2021
\(a,M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,M=1-\dfrac{1}{\sqrt{a}}< 1\\ c,a=3-2\sqrt{2}\Leftrightarrow\sqrt{a}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\\ \Leftrightarrow M=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}=\dfrac{-\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{2}\)
14 tháng 5 2023
\(A=\dfrac{2-\sqrt{a}-\sqrt{a}-3}{2\sqrt{a}+1}=-1\)
\(B=\dfrac{1}{1-\sqrt{2+\sqrt{3}}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}-1}-\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}+1}\)
\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)
\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)
4 tháng 12 2023
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)