1/2+1/4+1/6+1/8+1/16):x=3/1*2+3/2*3+3/3*4+...+3/15*16

\(\frac{53}{48}:x=3(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{15.16})\)

\(\frac{53}{48}:x=3(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{15}-\frac{1}{16})\)

\(\frac{53}{48}:x=3(1-\frac{1}{16})\)

\(\frac{53}{48}:x=\frac{45}{16}\)

\(x=\frac{53}{48}:\frac{45}{16}\)

\(x=\frac{53}{135}\)

Chúc bạn học tốt

A=1+2+2^2+2^3+....+2^9

2A=2+2^2+2^3+....+2^10

2A-A=2^10-1

A=2^10-1/2

B=5.2^8=(2^2+1).2^8=2^10+2^8

=>B>A

2A = 2(1 + 2 + 2² + .... + 2⁹ )

= 2 + 2² + 2³ + ..... + 2¹⁰

2A - A = (2 + 2² + 2³ + ..... + 2¹⁰) - (1 + 2 + 2² + .... + 2⁹ )

A = 2¹⁰ - 1  

B = 5.2⁸ = (2² + 1).2⁸ = 2¹⁰ + 2⁸

2¹⁰ - 1 < 2¹⁰ + 2⁸

=> A < B

\(\text{a) }\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\dfrac{3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ \\ =\dfrac{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{2^{32}-1}{3}\\ \)

\(\text{b) }24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right) \\ =\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^{16}-1\right)\left(5^{16}+1\right)\\ =5^{32}-1\\ \)

\(\text{c) }48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^{16}-1\right)\left(7^{16}+1\right)\\ =7^{32}-1\)

Đề là gì vậy bạn?

a)

Vì 2/9=6/27=8/36=12/54=16/72=18/81 nên:

2/9+6/27+8/36+12/54+16/72+18/81=

2/9+2/9+2/9+2/9+2/9+2/9=

2/9*6=

12/9=

4/3

Vậy 2/9+6/27+8/36+12/54+16/72+18/81=4/3

b)

Ta có:

1-2/5=3/5

1-2/7=5/7

1-2/9=7/9

...

1-2/99=97/99

Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=

3/5*5/7*7/9*...*97/99=

(3*5*7*...*97)/(5*7*9*...*99)=

3/99=

1/33

Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=1/33

c)

Gọi biểu thức 1/2+1/4+1/8+1/16+...+1/1024 là S,ta có:

S=1/2+1/4+1/8+1/16+...+1/1024

S*2=1+1/2+1/4+1/8+...+1/512

S*2-S=(1+1/2+1/4+1/8+...+1/512)-(1/2+1/4+1/8+1/16+...+1/1024)

S=1-1/1024

S=1023/1024

Vậy 1/2+1/4+1/8+1/16+...+1/1024=1023/1024

Cảm ơn bạn nhé!

\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-.............-\frac{1}{1024}\)

=> 2S = \(2x\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..........-\frac{1}{1024}\right)\)

     2S = \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..........-\frac{1}{512}\)

     2S - S = \(\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-........-\frac{1}{512}\right)\)- \(\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-........-\frac{1}{1024}\right)\)

=> S = \(1+\frac{1}{1024}=\frac{1024}{1024}+\frac{1}{1024}=\frac{1025}{1024}\)

Chắc chắn 100%

nhanh lên hộ mình vs

Đặt \(S=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(2S=2\times\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)\(=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{512}\)

\(2S-S=\left(1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)

\(\Rightarrow S=1+\frac{1}{1024}=\frac{1025}{1024}\)

\(\left(3\sqrt{7}\right)^2=63>28=\left(\sqrt{28}\right)^2\) hoặc \(3\sqrt{7}>2\sqrt{7}=\sqrt{28}\)

C1: $\sqrt{28}=\sqrt{4.7}=2\sqrt 7$

Ta có: $3>2$

$\Leftrightarrow 3\sqrt 7>3\sqrt 7$ hay $3\sqrt 7>\sqrt{28}$

C2: $3\sqrt{7}=\sqrt{63}$

Ta có: $63>28$

$\Leftrightarrow\sqrt{63}>\sqrt{28}$ hay $3\sqrt 7>\sqrt{28}$

a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

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

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

\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)