\(^3\)) và 2\(^6\)

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5 tháng 9 2018

*)\(2^3=8;2^6=64\)

\(8< 64=>2^3< 2^6\)

*)\(\left(\left(-\dfrac{1}{2}\right)^2\right)^3=\left(-\dfrac{1}{2}\right)^6=\left(-\dfrac{1^6}{2^6}\right)=\dfrac{1}{64}\)

\(\left(-\dfrac{1}{2}\right)^5=\left(\dfrac{-1^5}{2^5}\right)=\left(\dfrac{-1}{32}\right)\)

\(\dfrac{1}{64}>\left(\dfrac{-1}{32}\right)\)

\(=>\left(\left(-\dfrac{1}{2}\right)^2\right)^3>\left(-\dfrac{1}{2}\right)^5\)

AH
Akai Haruma
Giáo viên
9 tháng 7 2018

a) \(2^{2014}\)\(3^{1343}\)

Ta có:

\(2^{2014}=(2^3)^{\frac{2014}{3}}=8^{\frac{2014}{3}}< 9^{\frac{2014}{3}}\)

\(3^{1343}=(3^2)^{\frac{1343}{2}}=9^{\frac{1343}{2}}> 9^{\frac{2014}{3}}\)

\(\rightarrow 2^{2014}< 3^{1343}\)

b) \(31^{11}\)\(17^{44}\)

Có: \(17^{44}=(17^4)^{11}> (17.2)^{11}>31^{11}\)

AH
Akai Haruma
Giáo viên
9 tháng 7 2018

c)

\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)

\(\Rightarrow 2A=1+\frac{1}{2^1}+\frac{1}{2^2}+..+\frac{1}{2^{49}}\)

Lấy vế sau trừ vế trước thu được:

\(2A-A=1-\frac{1}{2^{50}}< 1\)

\(\Leftrightarrow A< 1\)

d) \(B=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(\Rightarrow 3B=1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

Lấy vế sau trừ vế trước:

\(\Rightarrow 3B-B=1-\frac{1}{3^{100}}< 1\)

\(\Leftrightarrow 2B< 1\Rightarrow B< \frac{1}{2}\)

\(A=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3-\dfrac{5}{2}+\dfrac{7}{3}\)

\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)

AH
Akai Haruma
Giáo viên
9 tháng 7 2018

b) \(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2018}\right)\)

\(=\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}....\frac{2018-1}{2018}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2017}{2018}=\frac{1.2.3...2017}{2.3.4...2018}=\frac{1}{2018}\)

c) Giữa các biểu thức là dấu nhân hay dấu cộng vậy bạn?

AH
Akai Haruma
Giáo viên
9 tháng 7 2018

d)

\(D=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(D=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

e) \(E=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)

\(2E=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(2E=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{99-97}{97.99}\)

\(2E=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

\(\Rightarrow E=\frac{16}{99}\)

19 tháng 11 2022

a: =>1/6x=-49/60

=>x=-49/60:1/6=-49/60*6=-49/10

b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2

=>x=17/15 hoặc x=-13/15

c: =>1,25-4/5x=-5

=>4/5x=1,25+5=6,25

=>x=125/16

d: =>2^x*17=544

=>2^x=32

=>x=5

i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5

=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2

=>x=14,4 hoặc x=9,6

j: =>(2x-1)(2x+1)=0

=>x=1/2 hoặc x=-1/2

8 tháng 8

17 tháng 11 2022

Bài 7:

x/1=z/2 nên x/6=z/12

=>x/6=y/9=z/12

=>x/2=y/3=z/4

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{27}{9}=3\)

=>x=6; y=9; z=12

25 tháng 3 2017

2) -12:\(\left(-\dfrac{5}{6}\right)^2\)=\(-12:\dfrac{25}{36}=-12\cdot\dfrac{36}{25}=-\dfrac{432}{25}\)

s) \(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)

= \(-\dfrac{1}{12}-\dfrac{55}{24}=-\dfrac{2}{24}-\dfrac{55}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)

t) \(-1,75-\left(-\dfrac{1}{9}-2\dfrac{1}{18}\right)=-1,75-\left(-\dfrac{2}{18}-\dfrac{37}{18}\right)\)

= -1,75-(\(-\dfrac{13}{6}\)) = \(-\dfrac{7}{4}+\dfrac{13}{6}=\dfrac{5}{12}\)

c) \(\left(\sqrt{\dfrac{1}{9}}-0,5\right)^3+\dfrac{-1}{3}=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^3-\dfrac{1}{3}\)

= \(\left(-\dfrac{1}{6}\right)^3-\dfrac{1}{3}=\dfrac{-1}{216}-\dfrac{1}{3}=-\dfrac{73}{216}\)

d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{4}{25}}\right)^2-2\dfrac{1}{2}=\left(\dfrac{1}{2}-\dfrac{2}{5}\right)^2-\dfrac{5}{2}\)

= \(\left(\dfrac{1}{10}\right)^2-\dfrac{5}{2}=\dfrac{1}{100}-\dfrac{250}{100}=-\dfrac{249}{100}=-2,49\)

Bài 1: 

a: \(=\dfrac{-1}{8}+1-\dfrac{9}{4}-1\)

\(=\dfrac{-1}{8}-\dfrac{18}{8}=\dfrac{-19}{8}\)

b: \(=4\cdot1-2\cdot\dfrac{1}{4}+3\cdot\dfrac{-1}{2}+1\)

\(=4-\dfrac{1}{2}-\dfrac{3}{2}+1\)

=5-2

=3

21 tháng 12 2018

ê

21 tháng 12 2018

2x-3y+5z=1 hoặc =-1

TH1: \(\dfrac{x}{y}\)=\(\dfrac{3}{2}\)=>\(\dfrac{x}{3}\)=\(\dfrac{y}{2}\)=>\(\dfrac{x}{15}\)=\(\dfrac{y}{10}\)

\(\dfrac{y}{z}\)=\(\dfrac{5}{7}\)=>\(\dfrac{y}{5}\)=\(\dfrac{z}{7}\)=>\(\dfrac{y}{10}\)=\(\dfrac{z}{14}\)

\(\Rightarrow\)\(\dfrac{x}{15}\)=\(\dfrac{y}{10}\)=\(\dfrac{z}{14}\)=>\(\dfrac{2x}{30}\)=\(\dfrac{3y}{30}\)=\(\dfrac{5z}{70}\)

Áp dụng tính chát dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x-3y+5z}{30-30+70}\)=\(\dfrac{1}{70}\)

=>x=1.15:7=\(\dfrac{3}{14}\)

y=\(\dfrac{1}{7}\)

z=\(\dfrac{1}{5}\)

TH2:............=-1 tự tính nhé làm tương tựvuimình còn phải ôn bài

26 tháng 6 2017

a, \(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{3}{7}\right)^2+2.\dfrac{3}{7}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)

\(=\dfrac{9}{49}+\dfrac{3}{7}+\dfrac{1}{4}=\dfrac{169}{196}\)

b, \(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2=\left(\dfrac{3}{4}\right)^2-2.\dfrac{3}{4}.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2\)

\(=\dfrac{9}{16}-\dfrac{5}{4}+\dfrac{25}{36}=\dfrac{1}{144}\)

c, \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.5^4.4^4}{5^{10}.4^5}=\dfrac{1}{5^2.4}=\dfrac{1}{100}\)

d, \(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4=\dfrac{\left(-10\right)^5}{3^5}.\dfrac{6^4}{5^4}\)

\(=\dfrac{5^5.\left(-2\right)^5.2^4.3^4}{3^5.5^4}=\dfrac{-\left(5.2^9\right)}{3}=\dfrac{-2560}{3}\)

Chúc bạn học tốt!!!