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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: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
`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 : \(a-3< 0\)
\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
Lời giải:
Đặt \((a^2-1)\sqrt{a^2-4}=m; a^3-3a-2=n\)
Ta thấy:
\(m^2=(a^2-1)^2(a^2-4)=(a-1)^2(a+2)(a+1)^2(a-2)\)
\(=(n+4)n\)
\(M=\frac{n+m}{n+4+m}=\frac{n+\sqrt{n(n+4)}}{n+4+\sqrt{n(n+4)}}\)
\(=\frac{\sqrt{n}(\sqrt{n}+\sqrt{n+4})}{\sqrt{n+4}(\sqrt{n+4}+\sqrt{n})}\)
\(=\sqrt{\frac{n}{n+4}}\)
\(C=\dfrac{a^2-3a-\left(a-1\right)\sqrt{a^2-4}+2}{a^2+3a-\left(a+1\right)\sqrt{a^2-4}+2}.\sqrt{\dfrac{a+2}{a-2}}\)
\(\Leftrightarrow\dfrac{a^2-3a+\left(-a+1\right)\sqrt{a^2-4}+2}{a^2+3a+\left(-a-1\right)\sqrt{a^2-4}+2}.\sqrt{\dfrac{a+2}{a-2}}\)
\(\Leftrightarrow\)\(\dfrac{\left[a^2-3a+\left(-a+1\right)\sqrt{a^2-4}+2\right]\sqrt{a+2}}{\left[a^2+3a+\left(-a-1\right)\sqrt{a^2-4}+2\right]\sqrt{a-2}}\)
\(\Leftrightarrow\)\(\dfrac{a^2\sqrt{a+2}-3a\sqrt{a+2}+\left(-a+1\right)\sqrt{a^2-4}\sqrt{a+2}+2\sqrt{a+2}}{a^2\sqrt{a-2}+3a\sqrt{a-2}+\left(-a-1\right)\sqrt{a^2-4}\sqrt{a-2}+2\sqrt{a-2}}\)
\(\Leftrightarrow\)\(\dfrac{a^2\sqrt{a+2}-3a\sqrt{a+2}+\left(-a+1\right)\sqrt{a^3+2a^2-4a-8}+2\sqrt{a+2}}{a^2\sqrt{a-2}+3a\sqrt{a-2}+\left(-a-1\right)\sqrt{a^3-2a^2-4a+8}+2\sqrt{a-2}}\)
a) \(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=-10\sqrt{2}+5.2-\left(18-30\sqrt{2}+25\right)\)
\(=-10\sqrt{2}+10-18+30\sqrt{2}-25\)
\(=20\sqrt{2}-33\)
b) câu b đề sai
\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)
\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)
\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)
\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)