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28 tháng 2 2016

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

Tử số bằng:

\(17^{36}+2.17^{18}+1-17^{36}-17^{17}-17^{19}-1\)

\(2.17^{18}-17^{17}-17^{19}\)

Vậy A - B < 0 => A < B

\(17^{17}\left(2.17-1-17^2\right)=-17^{17}.274<0\)

28 tháng 2 2016

Quản lí gì đâu làm hết mất không để người khác làm

18 tháng 3 2018

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 \(A< B\)

16 tháng 3 2018

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

17 tháng 5 2016

\(A=\frac{17^{18}+1}{17^{19}+1}\) 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 \(17^{19}+1< 17^{19}+10\Rightarrow1-A>1-B\)

                  \(\Rightarrow A< B\)

27 tháng 5 2019

áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{b+m}< 1\left(m\in N\right)\)

Ta có: \(A=\frac{17^{18}-1}{17^{20}-1}< \frac{17^{18}-1-16}{17^{20}-1-16}\)\(=\frac{17^{18}-17}{17^{20}-17}=\frac{17.\left(17^{17}-1\right)}{17.\left(17^{19}-1\right)}\)\(=\frac{17^{17}-1}{17^{19}-1}\)

\(\Rightarrow A< B\)

27 tháng 5 2019

\(A=\frac{17^{18}-1}{17^{20}-1}\Rightarrow17^2A=\frac{17^{18}-1}{17^{18}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}\left(1\right)\)

\(B=\frac{17^{17}-1}{17^{19}-1}\Rightarrow17^2B=\frac{17^{17}-1}{17^{17}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(2\right)\)

\(\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}< \frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\Rightarrow1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}>1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(3\right)\)

Từ \(\left(1\right);\left(2\right)\&\left(3\right)\Rightarrow17^2A>17^2B\Leftrightarrow A>B.\)

7 tháng 3 2018

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

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

Nhận thấy: 1719+1 > 1718+1  => \(\frac{16}{17^{18}+1}>\frac{16}{17^{19}+1}\)

=> 17B > 17A

=> B > A

12 tháng 5 2018

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

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

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

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

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

\(\Leftrightarrow17A< 17B\)

\(\Leftrightarrow A< B\)

12 tháng 5 2018

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 \(17^{19}+1>17^{18}+1\)

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

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

\(\Rightarrow B>A\)

11 tháng 3 2017

Bài 1:

Ta thấy A < 1

=> 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

Bài 2:

Ta thấy C < 1

=> C = \(\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)

Vậy C < D

14 tháng 3 2016

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

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

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