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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) \(E=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(2E=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(2E=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{99-97}{97.99}\)
\(2E=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow E=\frac{16}{99}\)
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
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 \(2^{505}\)
\(5^{202}=\left(5^2\right)^{101}=25^{101}\)
\(2^{505}=\left(2^5\right)^{101}=32^{101}\)
\(\Rightarrow25^{101}< 32^{101}\)
\(\Rightarrow5^{202}< 2^{505}\)
1) Tính
a) 253 : 52 = (52)3 : 52 = 56 : 52 = 54 = 625
\(b)\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^6=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^6=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^9\) d) 9 . 32 . \(\dfrac{1}{81}\) . 32 = 32 . 32 . \(\dfrac{1}{3^4}\) . 32 = 9
2) Tìm x thuộc Q, biết:
a) 3x + 2 = 27
=> 3x + 2 = 33
x + 2 = 3
x = 3 - 2
x = 1
b) \(\left(\dfrac{1}{2}x-3\right)^4=81\)
\(\Rightarrow\left(\dfrac{1}{2}x-3\right)^4=3^4\)
\(\dfrac{1}{2}x-3=3^{ }\)
\(\dfrac{1}{2}x=3+3\)
\(\dfrac{1}{2}x=9\)
\(x=9:\dfrac{1}{2}\)
\(x=18\)
c) \(\left(x-\dfrac{1}{2}\right)^3=-27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(-3\right)^3\)
\(x-\dfrac{1}{2}=-3\)
\(x=-3+\dfrac{1}{2}\)
\(x=\dfrac{-5}{2}\)
d) 5 . 5x + 1 = 125
5x + 1 = 125 : 5
5x + 1 = 25
5x + 1 = 52
x + 1 = 2
x = 2 - 1
x = 1.
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: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
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:
\(\Rightarrow 3B-B=1-\frac{1}{3^{100}}< 1\)
\(\Leftrightarrow 2B< 1\Rightarrow B< \frac{1}{2}\)