giúp mình vs

\(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\)

\(\Rightarrow4A=2^2\left(\dfrac{1}{2^2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{100}}\right)=1+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}\)

\(\Rightarrow3A=4A-A=1+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^2}-\dfrac{1}{2^4}-...-\dfrac{1}{2^{100}}=1-\dfrac{1}{2^{100}}\)

\(\Rightarrow A=\left(1-\dfrac{1}{2^{100}}\right):3=\dfrac{1}{3}-\dfrac{1}{2^{100}.3}< \dfrac{1}{3}\left(đpcm\right)\)

A=1/1^2+ 1/2^2+ 1/3^2+...+ 1/99^2+ 1/100^2

A=1+ 1/2^2+ 1/3^2+...+ 1/99^2+ 1/100^2

A<1+(1/2^2+1/2.3+1/3/4+...+1/98.99+1/99.100) (giữ nguyên phân số 1/2^2)

A<1+ (1/4+1/2-1/3+1/3-1/4+...+1/99-1/99+1/99-1/100)

A<1+ (1/4+1/2-1/100)

Mà 1/4+1/2-1/100 <1/4+1/2=3/4

=>A<1+3/4=7/4

x = 3- 1 - 1

x = 1

Vậy x =1

1 + 1 + x = 3

 2 + x = 3

 x = 3 - 2 

x = 1

Sửa đề: A=1/2^2+...+1/100^2

1/2^2<1/1*2

1/3^2<1/2*3

...

1/100^2<1/99*100

=>A<1-1/2+1/2-1/3+...+1/99-1/100

=>A<99/100<1

\(A=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

                                                               \(< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

                                                                \(< 1-\frac{1}{100}< 1\)

\(=>đpcm\)

Ủng hộ mk nha ^_-