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a) \(2^{2014}\) và \(3^{1343}\)
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) \(31^{11}\) và \(17^{44}\)
Có: \(17^{44}=(17^4)^{11}> (17.2)^{11}>31^{11}\)
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:
\(\Rightarrow 3B-B=1-\frac{1}{3^{100}}< 1\)
\(\Leftrightarrow 2B< 1\Rightarrow B< \frac{1}{2}\)
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) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{4}{25}}\right)^2-2\dfrac{1}{2}=\left(\dfrac{1}{2}-\dfrac{2}{5}\right)^2-\dfrac{5}{2}\)
= \(\left(\dfrac{1}{10}\right)^2-\dfrac{5}{2}=\dfrac{1}{100}-\dfrac{250}{100}=-\dfrac{249}{100}=-2,49\)
1,\(\dfrac{a}{b}=\dfrac{x}{y}\) khi ay=bx
2,
a,x=\(\dfrac{-1.12}{4}\)
x=\(\dfrac{-12}{4}=-3\)
b,\(\left(\dfrac{1}{3}\right)^{2x-1}=\left(\dfrac{1}{3}\right)^5\)
\(\Rightarrow\)2x-1=5
2x=6
x=6:2=3
c,\(\dfrac{4}{7}\).x=\(\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{3}{15}+\dfrac{10}{15}\)
\(\dfrac{4}{7}.x=\dfrac{13}{15}\)
\(x=\dfrac{13}{15}:\dfrac{4}{7}\)
x=\(\dfrac{13}{15}.\dfrac{7}{4}=\dfrac{91}{60}\)
3,ta có:\(5^{202}=\left(5^2\right)^{101}\)=\(25^{101}\)
2\(^{505}\)=\(\left(2^5\right)^{101}\)=\(32^{101}\)
vì 25<32 nên \(25^{101}< 32^{101}\) hay \(5^{202}< 2^{505}\)
1) \(\dfrac{a}{b}=\dfrac{x}{y}\) khi \(a.y=b.x\)
2) \(a,\dfrac{x}{12}=\dfrac{-1}{4}\)
\(\Rightarrow4x=-12\)
\(\Rightarrow x=-\dfrac{12}{4}=-3\)
Vậy x = -3
\(b,\left(\dfrac{1}{3}\right)^{2x-1}=\dfrac{1}{243}\)
\(\left(\dfrac{1}{3}\right)^{2x-1}=\left(\dfrac{1}{3}\right)^5\)
\(\Rightarrow2x-1=5\)
\(\Rightarrow x=\dfrac{5-1}{2}=2\)
Vậy x = 2
\(c,\dfrac{4}{7}x-\dfrac{2}{3}=\dfrac{1}{5}\)
\(\dfrac{4}{7}x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}x=\dfrac{13}{15}\)
\(\Rightarrow x=\dfrac{13}{15}:\dfrac{4}{7}=1\dfrac{31}{60}\)
Vậy \(x=1\dfrac{31}{60}\)
3) So sánh \(5^{202}\) và \(2^{505}\)
\(5^{202}=\left(5^2\right)^{101}=25^{101}\)
\(2^{505}=\left(2^5\right)^{101}=32^{101}\)
\(\Rightarrow25^{101}< 32^{101}\)
\(\Rightarrow5^{202}< 2^{505}\)
6:
\(4D=2^2+2^4+...+2^{202}\)
=>3D=2^202-1
hay \(D=\dfrac{2^{202}-1}{3}\)
7: \(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)
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
a/ \(19\dfrac{1}{3}.\dfrac{3}{7}-33\dfrac{1}{3}\)
\(=\dfrac{58}{3}.\dfrac{3}{7}-\dfrac{100}{3}\)
\(=\dfrac{58}{7}-\dfrac{100}{3}\)
\(=\dfrac{-526}{21}\)
b/ \(9.\left(\dfrac{-1}{2}\right)^2+\dfrac{1}{3}\)
\(=9.\dfrac{1}{4}+\dfrac{1}{3}\)
\(=\dfrac{9}{4}+\dfrac{1}{3}=\dfrac{31}{12}\)
c/ \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)
\(=\dfrac{61}{4}:\left(-\dfrac{5}{7}\right)-\dfrac{101}{4}:\left(-\dfrac{5}{7}\right)\)
\(=\left(-\dfrac{427}{20}\right)-\left(-\dfrac{707}{20}\right)\)
\(=14\)
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
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?
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) \(E=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(2E=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(2E=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{99-97}{97.99}\)
\(2E=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow E=\frac{16}{99}\)