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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) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{4}{25}}\right)^2-2\dfrac{1}{2}=\left(\dfrac{1}{2}-\dfrac{2}{5}\right)^2-\dfrac{5}{2}\)
= \(\left(\dfrac{1}{10}\right)^2-\dfrac{5}{2}=\dfrac{1}{100}-\dfrac{250}{100}=-\dfrac{249}{100}=-2,49\)
a: 2x(x-1/7)=0
=>x(x-1/7)=0
=>x=0 hoặc x=1/7
b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)
c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)
\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)
\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1\right\}\)
8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
\(S=\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\\ 25S=25\left(\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\right)\\ 25S=1+\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2016}}\\ 25S-S=\left(1+\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2016}}\right)-\left(\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\right)\\ 24S=1-\dfrac{1}{5^{2018}}< 1\\ \Rightarrow S< \dfrac{1}{24}\)
a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
\(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\Rightarrow4S=1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}+......-\dfrac{1}{2^{2002}}\Rightarrow4S+S=5S=1-\dfrac{1}{2^{2004}}< 1\Rightarrow S< 0,2\left(\text{đpcm}\right)\)
\(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(\dfrac{16}{5}:\dfrac{16}{5}+\dfrac{9}{2}\cdot\dfrac{76}{45}\right):\dfrac{-43}{2}\)
\(\Leftrightarrow-5< x< \left(1+\dfrac{38}{5}\right)\cdot\dfrac{-2}{43}=\dfrac{43}{5}\cdot\dfrac{-2}{43}=\dfrac{-2}{5}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1\right\}\)
1) \(25.\left(-\dfrac{1}{5}\right)^3+\dfrac{1}{5}-2.\left(-\dfrac{1}{2}\right)^2-\dfrac{1}{2}\)
\(=25.\left(-\dfrac{1}{125}\right)+\dfrac{1}{5}-2.\left(\dfrac{1}{4}\right)-\dfrac{1}{2}\)
\(=-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{2}-\dfrac{1}{2}\)
\(=-1\)
2) \(\dfrac{7}{8}:\left(\dfrac{2}{9}-\dfrac{1}{18}\right)+\dfrac{7}{8}:\left(\dfrac{1}{36}-\dfrac{5}{12}\right)\)
\(=\dfrac{7}{8}:\left(\dfrac{4}{18}-\dfrac{1}{18}\right)+\dfrac{7}{8}:\left(\dfrac{1}{36}-\dfrac{15}{36}\right)\)
\(=\dfrac{7}{8}:\dfrac{3}{18}+\dfrac{7}{8}:\left(-\dfrac{7}{18}\right)\)
\(=\dfrac{7}{8}:\left(\dfrac{3}{18}-\dfrac{7}{18}\right)\)
\(=\dfrac{7}{8}:\left(-\dfrac{2}{9}\right)\)
\(=\dfrac{7}{8}.\left(-\dfrac{9}{2}\right)\)
\(=-\dfrac{63}{16}\)
Ta có : S<\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{8.9}\)
S<\(\dfrac{2-1}{1.2}\)+\(\dfrac{3-2}{2.3}\)+...+\(\dfrac{9-8}{8.9}\)
S<1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+...+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
S<1-\(\dfrac{1}{9}\)
S<\(\dfrac{8}{9}\) (1)
Lại có : S>\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{9.10}\)
S>\(\dfrac{3-2}{2.3}\)+\(\dfrac{4-3}{3.4}\)+...+\(\dfrac{10-9}{9.10}\)
S>\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)
S>\(\dfrac{1}{2}\)-\(\dfrac{1}{10}\)
S>\(\dfrac{2}{5}\) (2)
Từ (1),(2) suy ra \(\dfrac{2}{5}\)<S<\(\dfrac{8}{9}\)