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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
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 ~
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}\)
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+17}{17^{19}+17}\)
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{17}+1}{17^{18}+1}=B\)
=> A < B
\(A=\frac{17^{18}+1}{17^{19}+1}\) <=> \(17A=\frac{17^{19}+17}{17^{19}+1}=\frac{17^{19}+1+16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)
\(B=\frac{17^{17}+1}{17^{18}+1}\)<=> \(17B=\frac{17^{18}+17}{17^{18}+1}=\frac{17^{18}+1+16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)
Nhận thấy: 1719+1 > 1718+1 => \(\frac{16}{17^{18}+1}>\frac{16}{17^{19}+1}\)
=> 17B > 17A
=> B > A
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
áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{b+m}< 1\left(m\in N\right)\)
Ta có: \(A=\frac{17^{18}-1}{17^{20}-1}< \frac{17^{18}-1-16}{17^{20}-1-16}\)\(=\frac{17^{18}-17}{17^{20}-17}=\frac{17.\left(17^{17}-1\right)}{17.\left(17^{19}-1\right)}\)\(=\frac{17^{17}-1}{17^{19}-1}\)
\(\Rightarrow A< B\)
\(A=\frac{17^{18}-1}{17^{20}-1}\Rightarrow17^2A=\frac{17^{18}-1}{17^{18}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}\left(1\right)\)
\(B=\frac{17^{17}-1}{17^{19}-1}\Rightarrow17^2B=\frac{17^{17}-1}{17^{17}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(2\right)\)
\(\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}< \frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\Rightarrow1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}>1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(3\right)\)
Từ \(\left(1\right);\left(2\right)\&\left(3\right)\Rightarrow17^2A>17^2B\Leftrightarrow A>B.\)
\(A=\frac{17^{18}-1}{17^{20}-1}\)
\(17^2A=\frac{17^2\left(17^{18}-1\right)}{17^{20}-1}=\frac{17^{20}-17^2}{17^{20}-1}=\frac{17^{20}-1-288}{17^{20}-1}=1-\frac{288}{17^{20}-1}\)
\(B=\frac{17^{17}-1}{17^{19}-1}\)
\(17^2B=\frac{17^2\left(17^{17}-1\right)}{17^{19}-1}=\frac{17^{19}-17^2}{17^{19}-1}=\frac{17^{19}-1-288}{17^{19}-1}=1-\frac{288}{17^{19}-1}\)
Ta có : \(\frac{288}{17^{20}-1}< \frac{288}{17^{19}-1}\)nên \(-\frac{288}{17^{20}-1}>-\frac{288}{17^{19}-1}\)
\(\Rightarrow A>B\)
NLA Saver sai rồi
* Xét : \(17^2A=\frac{17^2\left(17^{18}-1\right)}{17^{20}-1}=\frac{17^{20}-17^2}{17^{20}-1}=\frac{\left(17^{20}-1\right)+\left[\left(-17\right)^2+1\right]}{17^{20}-1}\)
\(=\frac{17^{20}-1}{17^{20}-1}+\frac{-17^{20}+1}{17^{20}-1}=1+\frac{-17^2+1}{17^{20}-1}=1+\frac{-288}{17^{20}-1}\)
* Tương tự xét \(17^2B=1+\frac{-288}{17^{19}-1}\)
Có : 1720 - 1 > 1719 - 1 => \(\frac{-288}{17^{20}-1}>\frac{-288}{17^{19}-1}\Rightarrow1+\frac{-288}{17^{20}-1}>1+\frac{-288}{17^{19}-1}\)
\(\hept{\begin{cases}17^2A>17^2B\\\text{mà}17^2>0\end{cases}}\Rightarrow A>B\)
ukm thanks bạn
Ta có:\(17^{18}-1< 17^{18}-1\)
=>\(\frac{17^{18}-1}{17^{20}-1}< \frac{17^{18}-1-16}{17^{20}-1-16}=\frac{17^{18}-17}{17^{20}-17}=\frac{17\left(17^{17}-1\right)}{17\left(17^{19}-1\right)}=\frac{17^{17}-1}{17^{19}-1}=B\)
=>A<B
Vậy A<B
chúc bạn học tốt !
nhớ k nha! thank!
=>
Ta có
\(\frac{a}{b}< 1=>\frac{a}{b}>\frac{a-n}{b-n}\)
\(\frac{17^{18}-1}{17^{20}-1}>\frac{17^{18}-17}{17^{20}-17}=\frac{17\left(17^{17}-1\right)}{17\left(17^{19}-1\right)}=B\)B
=>A>B
vậy ,........
hok tốt