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1 tháng 4 2017

a, Ta có : \(10^{15}\cdot11=10^{15}\left(10+1\right)=10^{16}+10^{15}\)

\(10^{16}+10^{15}>10^{16}+10\)

\(\Rightarrow\dfrac{10^{16}+10^{15}}{10^{16}+1}>\dfrac{10^{16}+10}{10^{16}+1}\)

Hay A>B

b, Ta có : \(C=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}}{10^{10}-1}+\dfrac{1}{10^{10}-1}\)

\(D=\dfrac{10^{10}-1}{10^{13}-3}=\dfrac{10^{10}}{10^{13}-3}+\dfrac{-1}{10^{13}-3}\)

\(\dfrac{10^{10}}{10^{10}-1}>\dfrac{10^{10}}{10^{13}-3};\dfrac{1}{10^{10}-1}>\dfrac{-1}{10^{13}-3}\)

\(\Rightarrow\dfrac{10^{10}+1}{10^{10}-1}>\dfrac{10^{10}-1}{10^{13}-3}\)

Hay C > D

26 tháng 9 2017

hehehahaleuleuhehehahaleuleuhehehahaleuleuhehehahaleuleuvhehehahaleuleuhehehahaleuleuhehehahaleuleuhehehahaleuleu

20 tháng 3 2018

Bệnh!!! oe

13 tháng 3 2018

a,A<B

b,A,<B

c,A<B

13 tháng 3 2018

a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)

Vậy A < B

b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)

\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)

Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)

c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:

 \(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)

Vậy A < B

10 tháng 5 2017

Ta có A = \(\dfrac{10^{15}-3-6}{10^{15}-3}\)= \(\dfrac{10^{15}-3}{10^{15}-3}-\dfrac{6}{10^{15}-3}=1-\dfrac{6}{10^{15}-3}\)

B = \(\dfrac{10^{16}-2-6}{10^{16}-2}=\dfrac{10^{16}-2}{10^{16}-2}-\dfrac{6}{10^{16}-2}\)= \(1-\dfrac{6}{10^{16}-2}\)

\(10^{15}-3\) = \(\overline{100...00}-3=\overline{9...7}\) (1)

\(10^{16}-2=\overline{100...000}-2=\overline{9...8}\) (2)

Từ (1) và (2) =>\(10^{15}-3< 10^{16}-2\) hay \(\dfrac{6}{10^{15}-3}>\dfrac{6}{10^{16}-2}\)

Vậy A > B

b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)

\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)

mà \(10^7-8< 10^8-7\)

nên A>B

c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)

\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)

mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)

nên A<B

6 tháng 4 2017

a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)

20 tháng 3 2017

d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B

cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A

Suy ra B>A(chuc ban hoc goi nhe)

5 tháng 5 2017

ta có: \(A=\dfrac{10.10^{10}-1}{10.10^{11}-1}=\dfrac{10^{10}-1}{10^{11}-1}\)

so sánh: \(A=\dfrac{10^{10}-1}{10^{11}-1}\)\(B=\dfrac{10^{10}+1}{10^{11}+1}\)

\(\Rightarrow A< B\)

5 tháng 5 2017

10A=\(\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)

10B =\(\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)

\(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{12}+1}\) =>\(1-\dfrac{9}{10^{12}-1}< 1+\dfrac{9}{10^{11}+1}\)

=> 10A < 10B

=> A<B

Vậy A < B