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LG :
x( 1 - 2 +2^2 - 2^3 ........+2^2006 - 2^ 2007) = 2^2008 - 1
co 1 - 2+ 2^2 - 2^3 .........- 2^2007 = - ( 2^2008 - 1) /3
Do đó x = -3
\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)
\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)
\(2A+A=2^{101}-2\)
\(A=\frac{2^{101}-2}{3}\)
b) tương tự
\(B=\frac{3^{101}+1}{4}\)
1) \(A=\left(2x^2+1\right)^4-3\ge0-3=-3\) (do \(\left(2x^2+1\right)^4\ge0\forall x\))
Dấu "=" xảy ra \(\Leftrightarrow\left(2x^2+1\right)=0\Leftrightarrow2x^2=-1\Leftrightarrow x^2=-\frac{1}{2}\) (vô lí)
Vậy đề sai ~v (hay là tui làm sai ta)
\(\frac{4}{7}=\frac{12}{21}\)
\(\Rightarrow\) \(x+4=12\Rightarrow x=8\)
\(\Rightarrow y+7=21\Rightarrow y=14\)
x + y = 8 + 14 = 22
****
suy ra (x + 4)7 = (y+7)4 mà x + y =22
7x+28 = 4y +28 suy ra x=22 -y (2)
7x = 4y (1)
từ (1) và (2) suy ra :7(22 - y)=4y
154 - 7y =4y
154 = 11y
suy ra y = 154 /11=14
x = 22-14=8
Ta co:\(B=\frac{2008}{1}+\frac{2007}{2}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\frac{2009-1}{1}+\frac{2009-2}{2}+...+\frac{2009-2007}{2007}+\frac{2009-2008}{2008}\)
\(B=\left(\frac{2009}{1}+\frac{2009}{2}+...+\frac{2009}{2008}\right)-\left(\frac{1}{1}+\frac{2}{2}+...+\frac{2008}{2008}\right)\)
\(B=2009+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)-2008\)
\(B=1+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2008}+\frac{1}{2009}\right)\)
Vay \(\frac{A}{B}=\frac{1}{2009}\)
A=2^100-2^99+2^98-2^97+..+2^2-2
=>2A=2^101-2^100+2^99-2^98+...+2^3-2^2
=>2A+A=(2^101-2^100+2^99-2^98+..+2^3-2^2)+(2^100-2^99+2^98-2^97+..+2^2-2)
=>3A=2^101-2
=>A=(2^101-2)/3
(2/101 - 2)/3 , tick nha