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18 tháng 1 2018

a) (-12).8 < (-19).3.

b) 11.(-2) > (-3).10.

c) (-16). 10 > (-32).11.

d) (-17).3 < (-22).2.

21 tháng 5 2017

d)

đặt A = 1 + 2 + 22 + ... + 280 

2A = 2 + 22 + 23 + ... + 281

2A - A = ( 2 + 22 + 23 + ... + 281 ) - ( 1 + 2 + 22 + ... + 280 )

A = 281 - 1 > 281 - 2

e) 

đặt \(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{899}{900}\)

\(A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{900}\right)\)

\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)

\(A=29-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)

đặt \(B=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\)

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}\)

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{29.30}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{29}-\frac{1}{30}\)

\(=1-\frac{1}{30}=\frac{29}{30}< 1\)

\(\Rightarrow A< 29\)

30 tháng 9 2020

So sánh C và D biết
C=1+13+13^2+...+13^13/1+13+13^2+...+13^12
D=1+11+11^2+...+11^13/1+11+11^2+...+11^12

14 tháng 4 2019

25 tháng 10 2017

\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)

\(B=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^{19}.3^9+3^{10}.2^{20}}\)

\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+3.2\right)}\)

\(B=\frac{7}{2.7}\)

\(B=\frac{1}{2}\)

\(C=\frac{2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}\)

\(C=\frac{2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)

\(C=\frac{2^{35}-2^{36}}{3^2.2^{34}}\)

\(C=\frac{2^{35}\left(1-2\right)}{3^2.2^{34}}\)

\(C=\frac{-2}{9}\)

\(D=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)

\(D=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^2.2^{20}}\)

\(D=\frac{2^{14}.2^8}{3.2^{23}-5.2^{22}}\)

\(D=\frac{2^{22}}{2^{22}\left(3.2-5\right)}\)

\(D=1\)

18 tháng 5 2021

\(a.\)

\(A=\)\(\frac{10^{15}+1}{10^{16}+1}\)

\(10A=\) \(\frac{10\left(10^{15}+1\right)}{10^{16}+1}\)

\(10A=\) \(\frac{10^{16}+10}{10^{16}+1}\)

\(10A=\)\(\frac{10^{16}+1+9}{10^{16}+1}\)

\(10A=\frac{10^{16}+1}{10^{16}+1}+\frac{9}{10^{16}+1}\)

\(10A=1+\frac{9}{10^{16}+1}\)

\(B=\frac{10^{16}+1}{10^{17}+1}\)

\(10B=\frac{10\left(10^{16}+1\right)}{10^{17}+1}\)

\(10B=\frac{10^{17}+10}{10^{17}+1}\)

\(10B=\frac{10^{17}+1+9}{10^{17}+1}\)

\(10B=\frac{10^{17}+1}{10^{17}+1}+\frac{9}{10^{17}+1}\)

\(10B=1+\frac{9}{10^{17}+1}\)

\(\Rightarrow10B< 10A\Rightarrow B< A\)\(\text{( vì tự làm ) }\)

19 tháng 5 2021

xin lỗi hôm qua mk đang làm thì phải đy học zoom học xong quên h mới nhơ ra làm típ :)

\(A=\frac{3}{8^3}+\frac{7}{8^4}=\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}\)

\(B=\frac{3}{8^4}+\frac{7}{8^3}=\frac{3}{8^4}+\frac{3}{8^3}+\frac{4}{8^3}\)

Vì \(\frac{4}{8^4}< \frac{4}{8^3}\)=.> A < B

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

Bài 6:

a: \(x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14-3}{21}=\dfrac{-17}{21}\)

d: \(x=\dfrac{9}{10}\cdot\dfrac{-5}{9}=\dfrac{-1}{2}\)

e: \(\Leftrightarrow x\cdot\dfrac{1}{3}=\dfrac{14}{21}-\dfrac{3}{21}=\dfrac{11}{21}\)

=>x=11/7

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)

\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)