\(A=\frac{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}}{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}}\)

Đặt tử số là B, mẫu số là C

\(B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

\(2B=2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)

\(2B-B=\left(2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)\)

\(B=2-\frac{1}{16}\)

\(B=\frac{32}{16}-\frac{1}{16}=\frac{31}{16}\)

\(C=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\)

\(2C=2-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\)

\(2C+C=\left(2-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\right)\)

\(3C=2+\frac{1}{16}\)

\(3C=\frac{32}{16}+\frac{1}{16}\)

\(3C=\frac{33}{16}\)

\(C=\frac{33}{16}:3=\frac{11}{16}\)

=> \(A=\frac{B}{C}=\frac{31}{16}:\frac{11}{16}=\frac{31}{16}.\frac{16}{11}=\frac{31}{11}\)

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(2\times A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)

\(A=1-\frac{1}{128}\)

\(A=\frac{127}{128}\)

\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

\(2\times B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)

\(B=1-\frac{1}{16}=\frac{15}{16}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Leftrightarrow4\times x+\frac{15}{16}=1\)

\(\Leftrightarrow4\times x=\frac{1}{16}\)

\(\Leftrightarrow x=\frac{1}{64}\)

1/2+1/4+1/8+1/16

=1/2+1/2-1/4+1/4-1/8+1/8-1/16

=1-1/16

=15/16

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)

\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^6}=1-\dfrac{1}{2^6}\)

Mọi người giúp mink nha

10,82896825 :V

\(\dfrac{63}{64}\)

\(\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{8}{64}+\dfrac{4}{64}+\dfrac{2}{64}+\dfrac{1}{64}=\dfrac{32+16+8+4+2+1}{64}=\dfrac{63}{64}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{18}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)

\(A=1-\dfrac{1}{64}\)

\(A=\dfrac{63}{64}\)

giúp em với ạ , em cấn gấp ,  4h em học rồi ạ . Cảm ơn những ai giúp em ạ , em tick luôn ạ

$4(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^8-1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^{16}-1)(3^{16}+1)$

$=\dfrac{3^{32}-1}{2}$

a: 4A=4+4^2+...+4^9

=>3A=4^9-1

=>A=(4^9-1)/3

b: 2A=1+1/2+...+1/2^7

=>A=1-1/256=255/256

c: =1-1/5+1/5-1/9+...+1/85-1/89

=1-1/89=88/89

d: =1/3(3/1*4+3/4*7+...+3/304*307)

=1/3(1-1/4+1/4-1/7+...+1/304-1/307)

=1/3*306/307=102/307

e: E=1-1/2+1/2-1/3+...+1/11-1/12

=1-1/12=11/12

g: =2/5(1-1/6+1/6-1/11+...+1/96-1/101)

=2/5*100/101=40/101