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\(B=10101\cdot\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\left(\dfrac{2}{222222}+\dfrac{5}{222222}\right)\)
\(=10101\cdot\dfrac{1}{31746}=\dfrac{7}{22}\)
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: \(=10101\cdot\left(\dfrac{10+5}{222222}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\dfrac{7}{222222}=\dfrac{7}{22}\)
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
bai....................kho......................wa.....................troi.........................thi.................lanh...............lai....................mua...................tich....................ung.....................ho..................minh....................nha.......................huhu
\(=3uv^2-\frac{1}{5}uv^2-367\frac{1}{4}uv^2-\frac{19}{5}uv^2+317\frac{1}{4}uv^2\)
\(=3uv^2-\frac{1}{5}uv^2-\frac{19}{5}uv^2=3uv^2-4uv^2=-uv^2\)
\(3uv^2-\left(\frac{1}{5}uv^2+367\frac{1}{4}uv^2\right)+\left(\frac{-19}{5}uv^2\right)+\left(367\frac{1}{4}uv^2\right)\)
=\(3uv^2-\frac{1}{5}uv^2-367\frac{1}{4}uv^2-\frac{19}{5}uv^2+317\frac{1}{4}uv^2\)
=\(3uv^2-\frac{1}{5}uv^2-\frac{19}{5}uv^2\)
=\(3uv^2-4uv^2\)
=\(-uv^2\)
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:
\(A=10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(=10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
\(=10101.\left(\frac{10}{222222}+\frac{5}{222222}-\frac{8}{222222}\right)\)
\(=10101.\left(\frac{10+5-8}{222222}\right)\)
\(=10101.\frac{7}{222222}\)
\(=\frac{7.10101}{22.10101}=\frac{7}{22}\)