Question

[kien le trong]

so sánh

 A= 17¹⁸ + 1 phần 17¹⁹ + 1

 và B = 17¹⁷ + 1 phần 17¹⁸ +1

Answer 1

[Kaito Kid]

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

Answer 2

[Huỳnh Quang Sang]

\(A=\frac{17^{18}+1}{17^{19}+1}\)

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 : \(A< B\)