cho \(a=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}\)
tính \(M=a^4-3a^3+2a+2015\)
toán 9
K
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7 tháng 7 2016
Bình phương a ta được
\(a^2=3+3+\sqrt{5+2\sqrt{3}}-\sqrt{5+2\sqrt{3}}+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}\)
\(a^2=6+2\sqrt{9-3\sqrt{5+2\sqrt{3}}+3\sqrt{5+2\sqrt{3}}-5-2\sqrt{3}}\)
\(a^2=6+2\sqrt{9-5-2\sqrt{3}}\Rightarrow a^2=6+2\sqrt{4-2\sqrt{3}}\Rightarrow a^2=6+2\sqrt{3+1-2.1.\sqrt{3}}\)\(a^2=6+2\sqrt{\left(\sqrt{3}-1\right)^2}\Rightarrow a^2=6+2\sqrt{3}-2=4+2\sqrt{3}=3+1+2.1.\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow a=\sqrt{3}+1\)
Rồi bạn tự thay vào tính típ nha
Chúc bạn học tốt
T I C K ủng hộ nha
25 tháng 9 2021
1) \(ĐK:3-2a>0\Leftrightarrow a< \dfrac{3}{2}\)
2) \(ĐK:2x-5< 0\Leftrightarrow x< \dfrac{5}{2}\)
3) \(ĐK:3-5a< 0\Leftrightarrow a>\dfrac{3}{5}\)
4) \(ĐK:a< 0\)
5) \(ĐK:-a\ge0\Leftrightarrow a\le0\)
28 tháng 6 2021
`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))`
28 tháng 6 2021
\(M=\left|3a-1\right|+2a-3\)
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 : \(a-3< 0\)
\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
3 tháng 8 2023
a: =(2căn 3-8căn 3)(căn 3-1)
=-6căn 3*(căn 3-1)
=-18+6căn 3
b: \(=\dfrac{6-2\sqrt{5}}{\sqrt{5}-3}-\sqrt{5}+2\)
=-2-căn 5+2=-căn 5
c: \(=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)
=\(3\sqrt{2a}-3a\cdot\sqrt{2a}\)
21 tháng 8 2021
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
b: \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{a}{2}\)
c: \(\sqrt{5a\cdot45a}-3a=-15a-3a=-18a\)
25 tháng 10 2020
a) Ta có: \(2\sqrt{3a}-\sqrt{12a^3}-5\cdot\sqrt{\frac{a}{3}}-\frac{1}{4}\cdot\sqrt{27a}\)
\(=2\sqrt{3a}-2a\sqrt{3a}-\frac{5\sqrt{a}}{\sqrt{3}}-\frac{1}{4}\cdot3\sqrt{3a}\)
\(=2\sqrt{3a}-\frac{3}{4}\sqrt{3a}-2a\sqrt{3a}-\frac{5\sqrt{a}}{\sqrt{3}}\)
\(=\frac{5}{4}\sqrt{3a}-2a\sqrt{3a}-5\sqrt{3a}\cdot\frac{1}{3}\)
\(=\frac{5}{4}\sqrt{3a}-\frac{5}{3}\sqrt{3a}-2a\sqrt{3a}\)
\(=\frac{-5}{12}\sqrt{3a}-2a\sqrt{3a}\)
b) Ta có: \(2a\sqrt{b+a}+\left(a+b\right)\cdot\sqrt{\frac{1}{a+b}}-\sqrt{a^3+a^2b}\)
\(=2a\sqrt{a+b}+\sqrt{\left(a+b\right)^2\cdot\frac{1}{a+b}}-a\sqrt{a+b}\)
\(=a\sqrt{a+b}+\sqrt{a+b}\)
\(=\left(a+1\right)\cdot\sqrt{a+b}\)
c) Ta có: \(2\sqrt{a}+5\sqrt{\frac{a}{9}}-a\sqrt{\frac{16}{a}}\cdot\sqrt{a^3}\)
\(=2\sqrt{a}+5\cdot\frac{\sqrt{a}}{3}-4a^2\)
\(=\frac{11}{3}\sqrt{a}-4a^2\)
2 tháng 7 2021
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: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)