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Q = (1 - \(\dfrac{\sqrt{a}-4a}{1-4a}\)) : \(\left[1-\dfrac{1+2a-2\sqrt{a}\left(2\sqrt{a}+1\right)}{1-4a}\right]\)
= \(\left(\dfrac{1-4a-\sqrt{a}+4a}{1-4a}\right):\left[\dfrac{1-4a-1-2a+4a+2\sqrt{a}}{1-4a}\right]\)
= \(\dfrac{1-\sqrt{a}}{1-4a}:\left(\dfrac{-2a+2\sqrt{a}}{1-4a}\right)\)
= \(\dfrac{1-\sqrt{a}}{1-4a}.\dfrac{1-4a}{2\sqrt{a}\left(1-\sqrt{a}\right)}\)
= \(\dfrac{1}{2\sqrt{a}}\) = \(\dfrac{\sqrt{a}}{2a}\)
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
a: ĐKXĐ: a>=0; b>=0; ab<>0; a<>1\(M=\dfrac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{1}{a-1}=\dfrac{1}{a-1}\)
b: M nguyên khi a-1 thuộc {1;-1}
=>a thuộc {2;0}
\(B=\left(\dfrac{1}{\sqrt[]{a}-1}-\dfrac{1}{\sqrt[]{a}}\right):\left(\dfrac{\sqrt[]{a}+1}{\sqrt[]{a}-2}-\dfrac{\sqrt[]{a}+2}{\sqrt[]{a}-1}\right)\left(1\right)\)
a) B xác định khi và chỉ khi :
\(\left\{{}\begin{matrix}a\ge0\\\sqrt[]{a}\ne0\\\sqrt[]{a}-1\ne0\\\sqrt[]{a}-2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)
b) \(\left(1\right)\Leftrightarrow B=\left(\dfrac{\sqrt[]{a}-\left(\sqrt[]{a}-1\right)}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{\left(\sqrt[]{a}+1\right)\left(\sqrt[]{a}-1\right)-\left(\sqrt[]{a}+2\right)\left(\sqrt[]{a}-2\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{a-1-\left(a-4\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{3}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)
\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right).\left(\dfrac{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}{3}\right)\)
\(\Leftrightarrow B=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}\)
`M=sqrt{(3a-1)^2}+2a-3`
`=|3a-1|+2a-3`
`=3a-1+2a-3(do \ a>=1/3)`
`=5a-4`
`N=sqrt{(4-a)^2}-a+5`
`=|4-a|-a+5`
`=a-4-a+5(do \ a>4)`
`=1`
`I=sqrt{(3-2a)^2}+2-7`
`=|3-2a|-5`
`=3-2a-5(do \ a<3/2)`
`=-2-2a`
`K=(a^2-9)/4*sqrt{4/(a-2)^2}`
`=(a^2-9)/4*|2/(a-2)|`
`=(a^2-9)/(2|a-2|)`
Nếu `3>a>2=>|a-2|=a-2`
`=>K=(a^2-9)/(2(a-2))`
Nếu `a<2=>|a-2|=2-a`
`=>K=(a^2-9)/(2(2-a))`
\(M=\left|3a-1\right|+2a-3\)
Mà \(a-\dfrac{1}{3}\ge0\)
\(\Rightarrow M=3a-1+2a-3=5a-4\)
\(N=\left|4-a\right|-a+5\)
Mà \(4-a< 0\)
\(\Rightarrow N=a-4-a+5=1\)
\(I=\left|3-2a\right|-5\)
Mà \(a-\dfrac{3}{2}< 0\)
\(\Rightarrow I=3-2a-5=-2a-2\)
K, Ta có : \(a-3< 0\)
\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
\(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(đk:a>0,a\ne1\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+2}=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{1}=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)
Để A nguyên
\(\Leftrightarrow A=\dfrac{\sqrt{a}-2}{\sqrt{a}}=1-\dfrac{2}{\sqrt{a}}\in Z\)
Do \(\sqrt{a}>0,\sqrt{a}\ne1\)
\(\Leftrightarrow\sqrt{a}\inƯ\left(2\right)=\left\{2\right\}\)
\(\Leftrightarrow a=4\)
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
a: \(A=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)
\(=\dfrac{\left(a-1\right)^2}{4a}\cdot\dfrac{-4\sqrt{a}}{a-1}\)
\(=\dfrac{-\left(a-1\right)}{\sqrt{a}}\)
b: \(=1+\left(\dfrac{\left(2\sqrt{a}-1\right)}{1-\sqrt{a}}+\dfrac{2a\sqrt{a}-\sqrt{a}+a}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}\)
Δ\(=1+\left(\dfrac{\left(-2\sqrt{a}+1\right)}{\sqrt{a}-1}+\dfrac{2a\sqrt{a}-\sqrt{a}+a}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}\)
\(=1+\left(\dfrac{-2a\sqrt{a}-\sqrt{a}+1+2a\sqrt{a}-\sqrt{a}+a}{a+\sqrt{a}+1}\cdot\dfrac{\sqrt{a}}{2\sqrt{a}-1}\right)\)
\(=1+\dfrac{\left(\sqrt{a}-1\right)^2\cdot\sqrt{a}}{\left(2\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{2a\sqrt{a}+2a+2\sqrt{a}-a-\sqrt{a}-1+a\sqrt{a}-2a+\sqrt{a}}{\left(2\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{3a\sqrt{a}-a+2\sqrt{a}-1}{\left(2\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)