Rút gọn biểu thức sau
A=1+1/2+1/2^2+1/2^3+...+1/2^2012
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Lời giải:
$(5x-1)^2+(5x+1)^2-2(1-25x^2)$
$=(5x-1)^2+(5x+1)^2-2(1-5x)(1+5x)$
$=(5x-1)^2+(5x+1)^2+2(5x-1)(5x+1)$
$=(5x-1+5x+1)^2$
$=(10x)^2=100x^2$
A=(ghi lại biieur thức)
2A=2+1+1/2+1/2^2+….+1/2^2011
2A-A=A=(2+1+1/2+1/2^2+….+1/2^2011)-(1+1/2+1/2^2+...+1/2^2012)
A=2-1/2^2012
1/2 A= 1/2+1/2^2+1/2^3+1/2^4+...........+1/2^2013
=>A-1/2A= 1 -1/2^2013
=>1/2A=1 -1/2^2013
=>A=(1 - 1/2^2013) : 1/2
2A = 2 + 1 + 1/2 + 1/22 + 1/23 + ... + 1/22011
mà A = 1 + 1/2 + 1/22 + 1/23 + ... + 1/22012
2A - A = 2 - 1/22012
A = 2 - 1/22012
Ta có A=1+1/2+1/2^2+1/2^3+........+1/2^2012
=>2A=2+1+1/2+1/2^2+.......+1/2^2011
=>2A-A=(2+1+1/2+1/2^2+.....+1/2^2011)-(1+1/2+1+1/2^2+1/2^3+.....+1/2^2012)
=>A=\(2-\frac{1}{2^{2012}}\)
\(A=\frac{2^{2013-1}}{2^{2012}}\)
A=đã cho.
1/2*A=1/2+1/2^2+1/2^3+...+1/2^2012+1/2^2013.
A-1/2*A=1-1/2^2013(khử).
1/2*A=1-1/2^2013.
A=2*(1-1/2^2013).
A=2-2/2^2013.
A=2-1/2^2012.
2A=2+1+1/2+1/2^2+1/2^3+...+1/2^2011
2A-A=(2+1+1/2+1/2^2+1/2^3+...+1/2^2011)-(1+1/2+1/2^2+1/2^3+...+1/2^2012)
A=2-2/2012
k cho mik nhé
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2013}}\)
\(\frac{1}{2}A-A=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2013}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(-\frac{1}{2}A=\frac{1}{2^{2013}}-1\)
\(A=\frac{\frac{1}{2^{2013}}-1}{2}\)
A = 1+1/2+1/2^2+1/2^3+.....+1/2^2012
2A= 2. (1+1/2+1/22+1/23+.....+1/22012)
2A= 2 + 1 + 1/2 + 1/22 + 1/23 + ...+ 1/22011
2A - A= (2 + 1 + 1/2 + 1/22 + 1/23+ ...+ 1/22011) - (1+1/2+1/22+1/23+.....+1/22012)
1A= 2 + 1 + 1/2 + 1/22 + 1/23 + ...+ 1/22011 - 1-1/2-1/22+1/23+.....+1/22012
1A= 2 - 1/22012
A= 2-1/22012
A= 2 - 1/22012
số mũ nữa nha
a: \(=\sqrt{3}-1-\sqrt{3}=-1\)
b: \(=2\sqrt{3}-10\sqrt{3}+4\sqrt{3}=-4\sqrt{3}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(A=2A-A=2-\frac{1}{2^{2011}}=\frac{2^{2012}-1}{2^{2011}}\)
Nhầm
\(A=2A-A=2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)