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10101 . \(\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
= 10101 . \(\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
= 10101 . \(\frac{7}{222222}\)
= \(\frac{7}{22}\)
Lời giải:
Ta có:
\(A=10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(=10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
\(=10101.\left(\frac{10}{222222}+\frac{5}{222222}-\frac{8}{222222}\right)\)
\(=10101.\left(\frac{10+5-8}{222222}\right)\)
\(=10101.\frac{7}{222222}\)
\(=\frac{7.10101}{22.10101}=\frac{7}{22}\)
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A = 1.2.3+23.1.2.3+33.1.2.3+43.1.2.3 /3.4.5+23.3.4.5+33.3.4.5+43.3.4.5
=1.2.3 (1+23+33+43)/3.4.5 (1+23+33+43)
=1.2.3/3.4.5
=1/10
B=10101 ( 15/333333 + 2/333333 +57 / 333333 )
B=10101 . 74 /333333
=74/33
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b: \(=8.2\left(11+\dfrac{94}{1591}-6-\dfrac{38}{1517}\right):\left(8+\dfrac{11}{43}\right)\)
\(=\dfrac{41}{5}\cdot\left(5+\dfrac{60}{1763}\right):\dfrac{355}{43}\)
\(=\dfrac{41}{5}\cdot\dfrac{8875}{1763}\cdot\dfrac{43}{355}\)
\(=5\)
c: \(=10101\cdot\left(\dfrac{10+5}{222222}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\dfrac{7}{222222}=\dfrac{7}{22}\)
\(\left(\frac{4}{9}+\frac{1}{3}\right)^2=\left(\frac{4}{9}+\frac{3}{9}\right)^2=\left(\frac{7}{9}\right)^2=\frac{49}{81}\)
\(\left(\frac{1}{2}-\frac{3}{5}\right)^3=\left(\frac{5}{10}-\frac{6}{10}\right)^3=\left(\frac{-1}{10}\right)^3=\frac{-1}{1000}\)
\(\left(\frac{-1}{5}\right)^5.\left(\frac{-6}{5}\right)^4=\frac{-5}{3125}.\frac{1296}{625}=\frac{-1296}{390625}\)
\(\left(\frac{3}{4}\right)^3:\left(\frac{3}{4}\right)^2:\left(-\frac{2}{5}\right)^3=\frac{3}{4}:\frac{-8}{125}=\frac{3}{4}.\frac{-125}{8}=\frac{-375}{32}\)
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a) \(\left|x\right|+\frac{1}{4}=\frac{1}{5}\)
\(\left|x\right|=\frac{1}{5}-\frac{1}{4}\)
\(\left|x\right|=\frac{-1}{20}\)(vô lý vì \(\left|x\right|\ge0\)với mọi x . Mà \(\frac{-1}{20}\)>0 )
Vậy không tồn tại x
b)\(\left|x+2\right|-\frac{1}{12}=\frac{1}{4}\)
\(\left|x+2\right|=\frac{1}{4}+\frac{1}{12}\)
\(\left|x+2\right|=\frac{1}{3}\)
\(\Rightarrow x+2\varepsilon\left\{\frac{1}{3};\frac{-1}{3}\right\}\)
+)\(x+2=\frac{1}{3}\Rightarrow x=\frac{-5}{3}\) +)\(x+2=\frac{-1}{3}\Rightarrow x=\frac{-7}{3}\)
Vậy \(x=\frac{-5}{3}\)hoặc \(x=\frac{-7}{3}\)
c)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)
\(\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)
\(\left|x+5\right|=\frac{-43}{42}\)( vô lý vì \(\left|x+5\right|\ge0\)với mọi x , mà \(\frac{-43}{42}< 0\))
Vậy không tồn tại x
d)\(\left|x+\frac{5}{6}\right|=\left|\frac{1}{5}-\frac{2}{3}\right|+\frac{-3}{4}\)
\(\left|x+\frac{5}{6}\right|=\frac{7}{15}+\frac{-3}{4}\)
\(\left|x+\frac{5}{6}\right|=\frac{-17}{60}\)( Vô lý vì \(\left|x+\frac{5}{6}\right|\ge0\)với mọi x mà \(\frac{-17}{60}< 0\))
Vậy không tồn tại x
\(B=10101\cdot\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\left(\dfrac{2}{222222}+\dfrac{5}{222222}\right)\)
\(=10101\cdot\dfrac{1}{31746}=\dfrac{7}{22}\)