Ta có A=17^18+1/17^19+1 < 17^18+1+16/17^19+1+16 = 17^18+17/17^19+17 = 17(17^17+1/17^18+1)= B
Vậy A<B
\(A=\frac{17^{18}+1}{17^{19}+1}\)
Ta có : \(17A=\frac{17(17^{18}+1)}{17^{19}+1}=\frac{17^{19}+17}{17^{19}+1}=\frac{17^{19}+1+16}{17^{19}+1}=1+\frac{17}{17^{19}+1}\) \((1)\)
\(B=\frac{17^{17}+1}{17^{18}+1}\)
Ta lại có : \(17B=\frac{17(17^{17}+1)}{17^{18}+1}=\frac{17^{18}+17}{17^{18}+1}=\frac{17^{18}+1+16}{17^{18}+1}=1+\frac{17}{17^{18}+1}\) \((2)\)
Từ 1 và 2 suy ra : \(1+\frac{16}{17^{19}+1}< 1+\frac{16}{17^{18}+1}\)
Nên \(17A< 17B\)
Hay \(A< B\)
Vậy : \(A< B\)
ta có \(A=\frac{17^{18}+1}{17^{19}+1}< 1\)
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+6}{17^{19}+1+6}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=B\)
\(< =>A< B\)
# T I C K MIK
A=\(\frac{17^{18}+1}{17^{19}+1}\)
\(\Rightarrow\)\(10A\)=\(\frac{17^{19}+1+16}{17^{19}+1}\)= \(1+\)\(\frac{16}{17^{19}+1}\)
\(\Rightarrow\)1\(10B\)=\(\frac{17^{18}+1+16}{17^{18}+1}\)=\(1+\frac{16}{17^{18}+1}\)=
\(\Rightarrow\)\(A\)>\(B\)
