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a: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà 17^19+1>17^18+1
nên A<B
b: \(2C=\dfrac{2^{2021}-2}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)
\(2D=\dfrac{2^{2022}-2}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)
2^2021-1<2^2022-1
=>1/2^2021-1>1/2^2022-1
=>-1/2^2021-1<-1/2^2022-1
=>C<D
3.13579/34567 = 40737/34567 = 34567+6170/34567
3.13580/34569 = 40740/34569 = 34569+6171/34569
vì : 34567+6170/34567 < 34569+6171/34569
nên: 3.13579/34567 < 3.13580/34569
vậy: 13579/34567 < 13580/34569
B=\(\dfrac{10^9+1}{10^{10}+1}< \dfrac{10^5+1+9}{10^{10}+1+9}=\dfrac{10^9+10}{10^{10}+10}=\dfrac{10.\left(10^8+1\right)}{10\left(10^9+1\right)}\)
= A
Ta có F < 1 nên
\(\dfrac{13580}{34569}< 1< \dfrac{13580+\left(-1\right)}{34569+\left(-1\right)}=\dfrac{13579}{34568}\)<\(\dfrac{13579}{34567}\)
Từ đó suy ra \(\dfrac{13580}{34569}< \dfrac{13579}{34567}\)hay\(\dfrac{13579}{34567}>\dfrac{13580}{34569}\)
Vậy E > F
2/
a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)
\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)
-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)
b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)
-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)
c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)
\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)
-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
Ta có: \(A=\dfrac{3^{10}+1}{3^9+1}\)
\(\Leftrightarrow A=\dfrac{3^{10}+3-2}{3^9+1}\)
hay \(A=3-\dfrac{2}{3^9+1}\)
Ta có: \(B=\dfrac{3^9+1}{3^8+1}\)
\(\Leftrightarrow B=\dfrac{3^9+3-2}{3^8+1}\)
hay \(B=3-\dfrac{2}{3^8+1}\)
Ta có: \(3^9+1>3^8+1\)
\(\Leftrightarrow\dfrac{2}{3^9+1}< \dfrac{2}{3^8+1}\)
\(\Leftrightarrow-\dfrac{2}{3^9+1}>-\dfrac{2}{3^8+1}\)
\(\Leftrightarrow-\dfrac{2}{3^9+1}+3>-\dfrac{2}{3^8+1}+3\)
hay A>B
Lời giải:
a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)
Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$
Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$
Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$
b) Rõ ràng $10^{11}-1< 10^{12}-1$.
Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$
Áp dụng kết quả phần a:
$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$
A = \(\dfrac{n^9+1}{n^{10}+1}\)
\(\dfrac{1}{A}\) = \(\dfrac{n^{10}+1}{n^9+1}\) = n - \(\dfrac{n-1}{n^9+1}\)
B = \(\dfrac{n^8+1}{n^9+1}\)
\(\dfrac{1}{B}\) = \(\dfrac{n^9+1}{n^8+1}\) = n - \(\dfrac{n-1}{n^8+1}\)
Vì n > 1 ⇒ n - 1> 0
\(\dfrac{n-1}{n^9+1}\) < \(\dfrac{n-1}{n^8+1}\)
⇒ n - \(\dfrac{n-1}{n^9+1}\) > n - \(\dfrac{n-1}{n^8+1}\)⇒ \(\dfrac{1}{A}>\dfrac{1}{B}\)
⇒ A < B
b: \(10A=\dfrac{10^9+10}{10^9+1}=1+\dfrac{9}{10^9+1}\)
\(10B=\dfrac{10^{10}+10}{10^{10}+1}=1+\dfrac{9}{10^{10}+1}\)
Vì \(10^9+1< 10^{10}+1\)
nên \(\dfrac{9}{10^9+1}>\dfrac{9}{10^{10}+1}\)
=>\(1+\dfrac{9}{10^9+1}>1+\dfrac{9}{10^{10}+1}\)
=>10A>10B
=>A>B