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a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)
Bài 1:
Ta có:
\(\left(a-b+c\right)^3=a^3-b^3+c^3-3a^2b+3a^2c+3ab^2+3b^2c+3ac^2-3bc^2-6abc\)
\(\Rightarrow\left(\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\right)^3=\frac{1}{9}-\frac{2}{9}+\frac{4}{9}-\frac{1}{3}.\sqrt[3]{2}+\frac{1}{3}.\sqrt[3]{4}+\frac{1}{3}.\sqrt[3]{4}+\frac{2}{3}.\sqrt[3]{2}\)
\(+\frac{2}{3}.\sqrt[3]{2}-\frac{2}{3}.\sqrt[3]{4}-\frac{4}{3}=\sqrt[3]{2}-1\)
\(\Rightarrow\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\)
\(2\sqrt{3a}-\sqrt{75a}+a\sqrt{\frac{6}{5}.\frac{5}{2a}}-\frac{2}{5}\sqrt{300a^3}\)
\(=2\sqrt{3a}-5\sqrt{3a}+a\sqrt{\frac{3}{2}}-\frac{2}{5}.10.a\sqrt{3a}\)
\(=-3\sqrt{3a}+\sqrt{\frac{3}{a}.a^2-4\sqrt{3a}}\)
\(=-3\sqrt{3a}+\sqrt{3a}-4a\sqrt{3a}\)
\(=-2\sqrt{3a}-4a\sqrt{3a}\)
\(=-2\sqrt{3a}\left(1+2a\right)\)
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: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
`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 : \(a-3< 0\)
\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)