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\(2\left(3x-2\right)-3\left(x-2\right)=-1\)
\(6x-4-3x+6=-1\)
\(3x+2=-1\)
\(3x=-1-2\)
\(3x=-3\)
\(x=-1\)
\(2\left(3-3x^2\right):3x\left(2x-1\right)=9\)
\(6-6x^2:6x^2-3x=9\)
\(6-x^2-3x=9\)
\(-x^2-3x+6=9\)
\(-x^2-3x=5\)
\(-x\left(x+3\right)=5\)
\(x=-5;x=2\)
2S1=2+2^2+2^3+2^4+..+2^64
2S1-S1=(2+2^2+2^3+2^4+..+2^64)-(1+2+2^2+2^3+...+2^63)
1S1=2^64-1
\(S=1+2+2^2+2^3+...+2^{63}\)
\(\Rightarrow2S=2+2^2+2^3+2^4+...+2^{64}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+2^4+...+2^{64}\right)-\left(1+2+2^2+2^3=...+2^{63}\right)\)
\(\Rightarrow S=2+2^2+2^3+2^4+...+2^{64}-1-2-2^2-2^3-...-2^{63}\)
\(\Rightarrow S=\left(2-2\right)+\left(2^2-2^2\right)+\left(2^3-2^3\right)+...+\left(2^{63}-2^{63}\right)+2^{64}-1\)
\(\Rightarrow S=2^{64}-1\)
b: \(A=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2003}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{2}{2003}\)
\(=2\left(\dfrac{1}{360}+\dfrac{1}{2003}\right)\)
\(B=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}\right)+\dfrac{5}{2003}\)
\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2003}\)
\(=5\left(\dfrac{1}{320}+\dfrac{1}{2003}\right)\)
Vì 1/360+1/2003<1/320+1/2003
nên A<B
B1 : S = 1 + 2 + 2^2 + 2^3 + ... + 2^2008 / 1 - 2^2009
Đặt A = 1 + 2 + 2^2 + 2^3 + ... + 2^2008
2A = 2 + 2^2 + 2^3 + 2^3 + 2^4 + ... + 2^2009
2A - A = ( 2 + 2^2 + 2^3 + 2^4 + ... + 2^2009 ) - ( 1 + 2 + 2^2 + 2^3 + ... + 2^2008 )
A = 2^2009 - 1
S = 2^2009 - 1 / 1 - 2^2009
S = -1
S=1+\(2^1\)+\(2^2\)+\(2^3\)+. . . .+\(2^{63}\)
2S=2+\(2^2\)+\(2^3\)+. . . .+\(2^{63}\)
2S-S=\(2^{64}\) - 1
S=\(2^{64}\)-1
ko có gì!