Question

[Hà Minh Đức]

So sánh

z=\(\frac{2^{100}+1}{2^{98}+1}\)                   và              t=\(\frac{2^{102}+1}{2^{100}+1}\)

Answer 1

[Phùng Minh Quân]

( ghi lại đề ) 

Ta có : 

\(\frac{1}{4}z=\frac{2^{100}+1}{2^{100}+4}=\frac{2^{100}+4-3}{2^{100}+4}=\frac{2^{100}+4}{2^{100}+4}-\frac{3}{2^{100}+4}=1-\frac{3}{2^{100}+4}\)

\(\frac{1}{4}t=\frac{2^{102}+1}{2^{102}+4}=\frac{2^{102}+4-3}{2^{102}+4}=\frac{2^{102}+4}{2^{102}+4}-\frac{3}{2^{102}+4}=1-\frac{3}{2^{102}+4}\)

Lại có : 

\(\frac{3}{2^{100}+4}>\frac{3}{2^{102}+4}\)

\(\Leftrightarrow\)\(-\frac{3}{2^{100}+4}< -\frac{3}{2^{102}+4}\)

\(\Leftrightarrow\)\(1-\frac{3}{2^{100}+4}< 1-\frac{3}{2^{102}+4}\)

\(\Leftrightarrow\)\(\frac{1}{4}z< \frac{1}{4}t\)

\(\Leftrightarrow\)\(z< t\)

Vậy \(z< t\)

Chúc bạn học tốt ~ 

Answer 2

[I don't need you]

ta có: \(T=\frac{2^{102}+1}{2^{100}+1}=\frac{2^2.\left(2^{100}+1\right)-3}{2^{100}+1}=\frac{2^2.\left(2^{100}+1\right)}{2^{100}+1}-\frac{3}{2^{100}+1}\)\(=4-\frac{3}{2^{100}+1}\)

\(Z=\frac{2^{100}+1}{2^{98}+1}=\frac{2^2.\left(2^{98}+1\right)-3}{2^{98}+1}=4-\frac{3}{2^{98}+1}\)

\(\Rightarrow\frac{3}{2^{100}+1}< \frac{3}{2^{98}+1}\)

\(\Rightarrow4-\frac{3}{2^{100}+1}>4-\frac{3}{2^{98}+1}\)

\(\Rightarrow T>Z\)