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

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

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

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

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

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

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

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

\(\Rightarrow17A< 17B\)

\(\Rightarrow A< B\)

Tham khảo :

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

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

Vì \(17^{19}>17^{18}=>17^{19}+1>17^{18}+1\)

\(=>\dfrac{16}{17^{19}+1}< \dfrac{16}{17^{18}+1}\)

\(=>17A< 17B=>A< B\)

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...

các bạn làm đúng hết r

17/18 - 1/6 bằng bao nhiêu

1) Phân tích A ra :

 A= 17¹⁷.17+$\frac{1}{17^{18}.17}$1‍1718.17 +1 So sánh với B ta có: A có 17¹⁸>17¹⁷ của B nhưng B lại có 1/17¹⁸>1/17¹⁹.

Mà 17¹⁸>1/17¹⁸ nên suy ra A>B

bn viết thế khó hiểu lắm

viết lại đi mik giải cho

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

\(\Leftrightarrow17A=\frac{17^{19}+17}{17^{19}+1}\)

\(\Leftrightarrow17A=\frac{17^{19}+1+16}{17^{19}+1}\)

\(\Leftrightarrow17A=\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}\)

\(\Leftrightarrow17A=1+\frac{16}{17^{19}+1}\)

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

\(\Leftrightarrow17B=\frac{17^{18}+17}{17^{18}+1}\)

\(\Leftrightarrow17B=\frac{17^{18}+1+16}{17^{18}+1}\)

\(\Leftrightarrow17B=\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}\)

\(\Leftrightarrow17B=1+\frac{16}{17^{18}+1}\)

Vì \(1+\frac{16}{17^{18}+1}>1+\frac{16}{17^{19}+1}\) nên 17B > 17A

Suy ra B > A

                            Giải

\(A=\frac{17^{18}+1}{17^{19}+1}\Leftrightarrow17A=\frac{17\left(17^{18}+1\right)}{17^{19}+1}\)

\(\Leftrightarrow17A=\frac{17^{19}+17}{17^{19}+1}\)

\(\Leftrightarrow17A=\frac{17^{19}+1+16}{17^{19}+1}\)

\(\Leftrightarrow17A=\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}\)

\(\Leftrightarrow17A=1+\frac{16}{17^{19}+1}\left(1\right)\)

\(B=\frac{17^{17}+1}{17^{18}+1}\Leftrightarrow17B=\frac{17\left(17^{17}+1\right)}{17^{18}+1}\)

\(\Leftrightarrow17B=\frac{17^{18}+17}{17^{18}+1}\)

\(\Leftrightarrow17B=\frac{17^{18}+1+16}{17^{18}+1}\)

\(\Leftrightarrow17B=\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}\)

\(\Leftrightarrow17B=1+\frac{16}{17^{18}+1}\left(2\right)\)

Từ (1) và (2) suy ra 17A < 17B

Suy ra A < B

ta có A=\(\frac{17^{18}+1}{17^{19}+1}\)<\(\frac{17^{18}+1+16}{17^{19}+1+16}\) (nếu a/b<1 thì a+c/b+c>a/b)

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

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

hay A<B

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

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

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

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

Do \(17^{19}+1< 17^{19}+10\Rightarrow1-A>1-B\)

                  \(\Rightarrow A< B\)