\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)

Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)

\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)

Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)

Đặt \(A=\left(1+x\right)\left(1+x^2\right)...\left(1+x^{32}\right)\)

\(\Rightarrow\left(1-x\right)A=\left(1-x\right)\left(1+x\right)\left(1+x^2\right)...\left(1+x^{32}\right)\)

\(=\left(1-x^2\right)\left(1+x^2\right)...\left(1+x^{32}\right)=\left(1-x^4\right)\left(1+x^4\right)...\left(1+x^{32}\right)\)

\(=\left(1-x^8\right)\left(1+x^8\right)...\left(1+x^{32}\right)=\left(1-x^{16}\right)\left(1+x^{16}\right)\left(1+x^{32}\right)\)

\(=\left(1-x^{32}\right)\left(1+x^{32}\right)=1-x^{64}\)

Vậy \(A=\frac{1-x^{64}}{1-x}\)

Sửa đề nhé:

Đặt \(B=1+x+x^2+...+x^{63}\)

\(\Rightarrow B\cdot x=x+x^2+x^3+...+x^{64}\)

\(\Rightarrow B-B\cdot x=1-x^{64}\)

\(\Rightarrow B\left(1-x\right)=1-x^{64}\)

\(\Rightarrow B=\frac{1-x^{64}}{1-x}\)

Vậy A = B ta có đpcm

Ta có :

\(S=\left(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}+\frac{127}{128}\right)-6\)

\(S=\left(\frac{64}{128}+\frac{102}{128}+\frac{112}{128}+\frac{120}{128}+\frac{124}{128}+\frac{126}{128}+\frac{127}{128}\right)-6\)

\(S=\frac{64+102+112+120+124+126+127}{128}-6\)

\(S=\frac{775}{128}-6\)

\(S=\frac{775}{128}-\frac{768}{128}\)

\(S=\frac{7}{128}\)

S=1/2+3/4+7/8+15/16+31/32+63/64+127/128 -6

S= 1-1/2 + 1-1/4 + 1-1/8 + 1-1/16 + 1-1/32 + 1-1/64+ 1-1/128 - 6

S= (1+1+1+1+1+1+1-6)- (1/2+1/4+1/8+1/16 + 1/32+1/64+1/128)

S= 1- 111/128

S= 17/128

(Làm lụi nha bn)

Ta có :

\(1+a+a^2+....+a^{63}\)

\(=\left(1+a\right)+a^2\left(1+a\right)+....+a^{62}\left(1+a\right)\)

\(=\left(1+a\right)\left(1+a^2+a^4+....+a^{62}\right)\)

\(=\left(1+a\right)\left[\left(1+a^2\right)+a^4\left(1+a^2\right)+.....+a^{60}\left(1+a^2\right)\right]\)

\(=\left(1+a\right)\left(1+a^2\right)\left(1+a^4+....+a^{60}\right)\)

.....

\(=\left(1+a\right)\left(1+a^2\right).....\left(1+a^{32}\right)\)

Có \(\left(1+a\right)\left(1+a^2\right)...\left(1+a^{32}\right)=\frac{\left(a-1\right)\left(a+1\right)\left(a^2+1\right)...\left(a^{32}+1\right)}{a-1}\)

\(=\frac{\left(a^2-1\right)\left(a^2+1\right)...\left(a^{32}+1\right)}{a-1}\)

\(...\)

\(=\frac{\left(a^{32}-1\right)\left(a^{32}+1\right)}{a-1}\)

\(=\frac{a^{64}-1}{a-1}\)

\(=\frac{\left(a-1\right)\left(a^{63}+a^{62}+...+a^2+a+1\right)}{a-1}\)

\(=a^{63}+a^{62}+...+a^2+a+1\)

Vậy ...

ta có (a-1)(1+a+a²+......+a⁶³)=a⁶⁴-1

        (a-1)(a+1)(a²+1)....(a³²+1)=a⁶⁴-1

=159 nha

sai đề rùi

Q=\(\dfrac{1}{2}+\left(\dfrac{3}{4}+\dfrac{7}{8}\right)+\left(\dfrac{15}{16}+\dfrac{31}{32}\right)+\left(\dfrac{63}{64}+\dfrac{127}{128}\right)-6\)

Q=\(\dfrac{1}{2}+\dfrac{13}{8}+\dfrac{61}{32}+\dfrac{253}{128}\)\(-6\)

Q= \(\dfrac{64}{128}+\dfrac{208}{128}+\dfrac{244}{128}+\dfrac{253}{128}-6\)

Q= \(\dfrac{769}{128}-6\)

Q=\(\dfrac{769}{128}-\dfrac{768}{128}\)

Q= \(\dfrac{1}{128}\)