Question

So sánh A=\(=\frac{4^{15}+1}{4^{17}+1}\) và B=\(\frac{4^{12}+1}{4^{14}+1}\)

Answer 1

* Cách 1 : 

Ta có : 

\(16A=\frac{4^{17}+16}{4^{17}+1}=\frac{4^{17}+1+15}{4^{17}+1}=\frac{4^{17}+1}{4^{17}+1}+\frac{15}{4^{17}+1}=1+\frac{15}{4^{17}+1}\)

\(16B=\frac{4^{14}+16}{4^{14}+1}=\frac{4^{14}+1+15}{4^{14}+1}=\frac{4^{14}+1}{4^{14}+1}+\frac{15}{4^{14}+1}=1+\frac{15}{4^{14}+1}\)

Vì \(\frac{15}{4^{17}+1}< \frac{15}{4^{14}+1}\) nên \(1+\frac{15}{4^{17}+1}< 1+\frac{15}{4^{14}+1}\)

\(\Rightarrow\)\(16A< 16B\) hay \(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

Answer 2

\(4^2.A=\frac{4^2\left(4^{15}+1\right)}{4^{17}+1}\); \(4^2.B=\frac{4^2\left(4^{12}+1\right)}{4^{14}+1}\)

=> \(4^2.A=\frac{4^{17}+4^2}{4^{17}+1}\);\(4^2.B=\frac{4^{14}+4^2}{4^{14}+1}\)

=> \(4^2.A=\frac{4^{17}+1+4^2-1}{4^{17}+1}\); \(4^2.B=\frac{4^{14}+1+4^2-1}{4^{14}+1}\)

=> \(4^2.A=\frac{4^{17}+1}{4^{17}+1}+\frac{4^2-1}{4^{17}+1}\); \(4^2.B=\frac{4^{14}+1}{4^{14}+1}+\frac{4^2-1}{4^{14}+1}\)

=> \(4^2.A=1+\frac{4^2-1}{4^{17}+1}\); \(4^2.B=1+\frac{4^2-1}{4^{14}+1}\)

Mà \(4^{17}>4^{14}\)

=> \(4^{17}+1>4^{14}+1\)

=> \(\frac{4^2-1}{4^{17}+1}< \frac{4^2-1}{4^{14}+1}\)

=> \(1+\frac{4^2-1}{4^{17}+1}< 1+\frac{4^2-1}{4^{14}+1}\)

=> \(4^2.A< 4^2.B\)

=> \(A< B\)

Answer 3

* Cách 2 : 

Ta có công thức : 

\(\frac{a}{b}< \frac{a+m}{b+m}\) \(\left(\frac{a}{b}< 1;a,b,m\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(A=\frac{4^{15}+1}{4^{17}+1}< \frac{4^{15}+1+63}{4^{17}+1+63}=\frac{4^{15}+64}{4^{17}+64}=\frac{4^{15}+4^3}{4^{17}+4^3}=\frac{4^3\left(4^{12}+1\right)}{4^3\left(4^{14}+1\right)}=\frac{4^{12}+1}{4^{14}+1}=B\)

\(\Rightarrow\)\(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~