Question

(\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)).x=\(\frac{2^{100}-1}{2^{100}}.100\)

Answer 1

Tìm x à?

Answer 2

trông loạn não ghê chờ tí

Answer 3

Đặt A=VT ta có:

\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

\(2A=1+\frac{1}{2}+...+\frac{1}{2^{100}}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{200}}=\frac{2^{200}-1}{2^{200}}\)

Thay A vào VT ta được:\(\frac{2^{200}-1}{2^{200}}x=\frac{2^{100}-1}{2^{100}}.100\)

=>x=100

Answer 4

Đặt S = 1/2+1/2²+1/2³+...+1/2¹⁰⁰

=> 1/2.S= 1/4+1/2³+1/2⁴+...+1/2¹⁰¹

=> S - 1/2S = 1/2.S= (1/4+1/2³+1/2⁴+...+1/2¹⁰¹ )-(1/2+1/2²+1/2³+...+1/2¹⁰⁰)

=> 1/2.S=(1/4+1/2¹⁰¹)-1/2

=> S= (1/4+1/2¹⁰¹)-1/2 : 1/2

rồi bây h thì oki

Answer 5

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

Thay A vào giả thiết của đề bài,ta được:

\(\left(1-\frac{1}{2^{100}}\right).x=\frac{2^{100-1}}{2^{100}}.100\)

\(\Rightarrow\left(\frac{2^{100}-1}{2^{100}}\right).x=\frac{2^{100}-1}{2^{100}}.100\)

\(\Rightarrow x=100\)

 

Answer 6

Ôg Thắng nhanh v~~~

Answer 7

Đặt A=VT ta có:

$2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)$2A=2(12 +122 +...+12100 )

$2A=1+\frac{1}{2}+...+\frac{1}{2^{100}}$2A=1+12 +...+12100 

$2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)$2A−A=(1+12 +...+12100 )−(12 +122 +...+12100 )

$A=1-\frac{1}{2^{200}}=\frac{2^{200}-1}{2^{200}}$A=1−12200 =2200−12200 

Thay A vào VT ta được:$\frac{2^{200}-1}{2^{200}}x=\frac{2^{100}-1}{2^{100}}.100$2200−12200 x=2100−12100 .100

=>x=100

 

Answer 8

Thiên Ngoại Phi Tiên : ko làm đc thì xéo,đừng có copy

Answer 9

Đặt S = 1/2+1/2²+1/2³+...+1/2¹⁰⁰

=> 1/2.S= 1/4+1/2³+1/2⁴+...+1/2¹⁰¹

=> S - 1/2S = 1/2.S= (1/4+1/2³+1/2⁴+...+1/2¹⁰¹ )-(1/2+1/2²+1/2³+...+1/2¹⁰⁰)

=> 1/2.S=(1/4+1/2¹⁰¹)-1/2

=> S= (1/4+1/2¹⁰¹)-1/2 : 1/2