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

\(17A=\frac{17^{19}+17}{17^{19}+1}=\frac{\left(17^{19}+1\right)+16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)

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

\(17B=\frac{17^{18}+17}{17^{18}+1}=\frac{\left(17^{18}+1\right)+16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

\(\text{Vì}\)\(1+\frac{16}{17^{19}+1}< 1+\frac{16}{17^{18}+1}\)

\(\Leftrightarrow17A< 17B\)

\(\Leftrightarrow A< B\)

Trả lời

\(17A=\frac{\left(17^{18}+1\right)17}{17^{19}+1}=\frac{17^{19}+17}{17^{19}+1}=\frac{17^{19}+1+16}{17^{19}+1}=\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)

\(17B=\frac{\left(17^{17}+1\right)17}{17^{18}+1}=\frac{17^{18}+17}{17^{18}+1}=\frac{17^{18}+1+16}{17^{18}+1}=\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

Vì \(17^{19}+1>17^{18}+1\)

\(\Rightarrow\frac{16}{17^{18}+1}>\frac{16}{17^{19}+1}\)

\(\Rightarrow1+\frac{16}{17^{18}+1}>1+\frac{16}{17^{19}+1}\)

\(\Rightarrow B>A\)

bn viết thế khó hiểu lắm

viết lại đi mik giải cho

Tham khảo :

cảm ơn chị nhiều.

Nếu nghĩ kĩ thì thấy bài này cũng đơn giản thôi.Thử xem cách giải của mk nè:

Giải: Ta có: A=\(\frac{17^{18}+1}{17^{19}+1}\)                                                        B=\(\frac{17^{17}+1}{17^{18}+1}\)

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

             17A=\(\frac{\left(17^{19}+1\right)+16}{17^{19}+1}\)                                       17B=\(\frac{\left(17^{18}+1\right)+16}{17^{18}+1}\)

               17A=\(\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}\)                             17B=\(\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}\)

               17A=\(1+\frac{16}{17^{19}+1}\)                                            17B= \(1+\frac{16}{17^{18}+1}\)

 Lại có: 17¹⁹+1>17¹⁸+1

 Suy ra:\(\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)

             17A<17B

             A<B

Vậy A<B

\(\text{Ta có:}\frac{17^{18}+1}{17^{19}+1}\)

\(\Rightarrow17A=\frac{17^{19}+1+16}{17^{19}+1}\)

\(\Rightarrow17A=1+\frac{16}{17^{19}+1}\)

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

\(\Rightarrow17B=\frac{17^{18}+1+16}{17^{18}+1}\)

\(\Rightarrow17B=1+\frac{16}{17^{18}+1}\)

\(\text{Vì }\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)

\(\Rightarrow17A< 17B\)

\(\Rightarrow A< B\)

Ta có công thức :

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

Áp dụng vào ta có : 

\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}\)

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

\(=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}\)

\(\Leftrightarrow\frac{17^{17}+1}{17^{18}+1}\)'

\(\Rightarrow=B\)

Vậy \(A< B\)

17/18 - 1/6 bằng bao nhiêu

Ta có:

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

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

\(17A=\frac{(17^{19}+1)+16}{(17^{19}+1)}=1+\frac{16}{17^{19}+1}\)          (1)

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

\(17B=\frac{17\left(17^{17}+1\right)}{17^{18}+1}=\frac{17^{18}+17}{17^{18}+1}\)

\(17B=\frac{(17^{18}+1)+16}{(17^{18}+1)}=1+\frac{16}{17^{18}+1}\)          (2)

Từ (1) và (2) => \(1+\frac{16}{17^{19}+1}< 1+\frac{16}{17^{18}+1}\)

=>\(17A< 17B\)

Hay \(A< B\)

Vậy \(A< B\)

Ta có công thức : 

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

Áp dụng vào ta có : 

\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

Ta có: \(A=\frac{17^{18}+1}{17^{19}+1}

Để so sánh A =17¹⁸+1/17¹⁹⁺¹ và B=17¹⁷+1/17¹⁸+1

=>Ta xét bài toán phụ sau

a/b<1 thì a/b<a+/b+m

vì a/b<1=>a<b mà m thuộc N*

=>a.m<b.m=>ab+am<ab+bm

a/b=a.(b+m0/b.(b+m)/b(b+m=ab+am/b(b+m)<ab+bm/b(b+m)

Vì b(b+m)>0=>a/b<ab+bm/b(b+m)=b(a+m)/b(b+m)=a+m/b+m

=>.a/b<a+m/b+m(1)

vì 17¹⁸+ 1 < 17¹⁹+1

=>A<1

(1)=>17¹⁸+1/17¹⁹+1<17¹⁸+1+16/17¹⁹+1+16

<=>A<17¹⁷+17/17¹⁹+17=17(17¹⁷+1)/17917¹⁸+1)

<=>A<17¹⁷+1/17¹⁸+1=B

<=>A<B

Vậy...