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a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012
2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013
3M=2^0+2^2013
M=(2^0+2^2013)÷3
Vậy.......
b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012
3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013
4N=3-3^2013
N=(3-3^2013)÷4
Vậy........
K tao nhé ko lên lớp tao đánh m😈😈😈
`a)đk:a>0,a ne 9`
`A=((sqrta+3+sqrta-3)/(a-9)).((sqrta-3)/sqrta)`
`=((2sqrtx)/(a-9)).((sqrta-3)/sqrta)`
`=2/(sqrta+3)`
`b)A>1/2`
`<=>2/(sqrta+3)>1/2`
`<=>sqrta+3<4`
`<=>sqrta<1`
`<=>a<1`
KẾt hợp đkxđ:`0<x<1`
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne9\end{matrix}\right.\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-3}+\dfrac{1}{\sqrt{a}+3}\right)\left(1-\dfrac{3}{\sqrt{a}}\right)\)
\(=\dfrac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\cdot\dfrac{\sqrt{a}-3}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}+3}\cdot\dfrac{1}{\sqrt{a}}\)
\(=\dfrac{2}{\sqrt{a}+3}\)
b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{2}{\sqrt{a}+3}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{4-\left(\sqrt{a}+3\right)}{2\left(\sqrt{a}+3\right)}>0\)
mà \(2\left(\sqrt{a}+3\right)>0\forall a\)
nên \(1-\sqrt{a}>0\)
\(\Leftrightarrow\sqrt{a}< 1\)
hay a<1
Kết hợp ĐKXĐ, ta được: 0<a<1
a, A = 7 - 4 3 + 1 2 - 3 = 2 - 3 + 2 + 3 = 4
b, B = sin 2 19 0 + cos 2 19 0 + tan 19 0 - c o t 71 0
= sin 2 19 0 + cos 2 19 0 + tan 19 0 - tan 19 0 = 1
\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}-1}+1}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}-1}-1\right)}{\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1}+2\right)}{\sqrt{3}}=\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1}+2\)
a) \(A=1+3+3^2+...+3^{100}\)
\(3A=3+3^2+3^3+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...-2^3+2^2-2+1\)
\(2B=2^{101}-2^{100}+2^{99}-2^{98}+...-2^4+2^3-2^2+2\)
\(B+2B=\left(2^{100}-2^{99}+...-2+1\right)+\left(2^{101}-2^{100}+...-2^2+2\right)\)
\(3B=2^{101}+1\)
\(B=\frac{2^{101}+1}{3}\)