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a,\(\frac{31}{1000}\)
b,\(\frac{8}{108}\)
c,0
a,\(\frac{49}{97}\)
b,\(\frac{-1}{4751}\)
Bài 1:
\(\Leftrightarrow-\dfrac{5}{7}:x=-\dfrac{7}{18}-\dfrac{1}{6}=\dfrac{-7}{18}-\dfrac{3}{18}=\dfrac{-10}{18}=\dfrac{-5}{9}\)
=>x=5/9:5/7=7/9
Bài 2:
a: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{1000}{999}=\dfrac{1000}{2}=500\)
b: \(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-999}{1000}\)
\(=-\dfrac{1}{1000}\)
Câu c:
C = 3/2^2.8/3^2.15/4^2...99/10^2
C = \(\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\ldots\times\) \(\frac{9.11}{10.10}\)
C = \(\frac{1.2.3\ldots9}{2.3\ldots10}\) x \(\frac{3.4.\ldots11}{2.3.\ldots10}\)
C = \(\frac{1}{10}\) x \(\frac{11}{2}\)
C = 11/20
Bài 1:
1/6 + -5/7 : x = - 7/18
5/7: x = 1/6 + 7/18
5/7: x = 3/18 + 7/18
5/7: x = 5/9
x = 5/7 : 5/9
x = 5/7 x 9/5
x = 9/7
Vậy x = 9/7
Bài 2:
Câu a:
(1/2 + 1).(1/3 + 1).(1/4 + 1)...(1/999 + 1)
= (1/2 + 2/2).(1/3 + 3/3)...(1/999 + 999/999)
= 3/2.4/3....1000/999
= 1000/2
= 500
Câu a:
(4 5/37 - 3 4/5 + 8 15/29) - (3 5/37 - 6 14/29)
= (4 5/37 - 3 5/37) + (8 15/29 + 6 14/29) - 3 4/5
= 1 + 15 - 3 4/5
= 16 - 3 - 4/5
= 13 - 4/5
= 61/5
Câu b:
(2\(\frac56\) + 1\(\frac49\)) : (10\(\frac{1}{12}\) - 9\(\frac12\))
= (17/6 + 13/9) :(121/12 - 19/2)
= (51/18 + 26/18) :(121/12 - 114/2)
= 77/18 : 7/12
= 77/18 x 12/7
= 22/3
\(a;\dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{2}{3}:\dfrac{3}{2}\\ \dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{4}{9}\\ \dfrac{3}{2}x=\dfrac{4}{9}+\dfrac{2}{3}=\dfrac{10}{9}\\ x=\dfrac{10}{9}:\dfrac{3}{2}=\dfrac{20}{27}\\ b;\left(\dfrac{9}{11}-x\right):\left(-\dfrac{10}{11}\right)=1-\dfrac{4}{5}\\ \left(\dfrac{9}{11}-x\right):\left(-\dfrac{10}{11}\right)=\dfrac{1}{5}\\ \dfrac{9}{11}-x=\dfrac{1}{5}\cdot\left(-\dfrac{10}{11}\right)\\ \dfrac{9}{11}-x=-\dfrac{2}{11}\\ x=\dfrac{9}{11}-\left(-\dfrac{2}{11}\right)=\dfrac{9}{11}+\dfrac{2}{11}\\ x=1\\ c;-\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\\ -\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}\\ -\dfrac{11}{12}x=-\dfrac{11}{12}\\ x=\left(-\dfrac{11}{12}\right):\left(-\dfrac{11}{12}\right)=1\)
\(d;-\dfrac{5}{4}-\left(1\dfrac{1}{2}+x\right)=4,5\\ \Leftrightarrow-\dfrac{5}{4}-\left(\dfrac{3}{2}+x\right)=4,5\\\dfrac{3}{2}+x=-\dfrac{5}{4}-4,5\\ \dfrac{3}{2}+x=-\dfrac{23}{4}\\ x=-\dfrac{23}{4}-\dfrac{3}{2}\\ x=-\dfrac{29}{4}\\ đ;\left(\dfrac{3}{4}-x:\dfrac{2}{15}\right)\cdot\dfrac{1}{5}=-2,6\\ \dfrac{3}{4}-x:\dfrac{2}{15}=-2,6:\dfrac{1}{5}\\ \dfrac{3}{4}-x:\dfrac{2}{15}=-13\\ x:\dfrac{2}{15}=\dfrac{3}{4}-\left(-13\right)\\ x:\dfrac{2}{15}=\dfrac{55}{4}\\ x=\dfrac{55}{4}\cdot\dfrac{2}{15}=\dfrac{11}{6}\\ e;3-\left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=3-\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{7}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{3}:\dfrac{2}{3}=\dfrac{7}{2}\\ x=\dfrac{1}{6}-\dfrac{7}{2}=-\dfrac{10}{3}\)
\(f;\left(1-2x\right)\cdot\dfrac{4}{5}=\left(-2\right)^3\\ \left(1-2x\right)\cdot\dfrac{4}{5}=-8\\ 1-2x=-8:\dfrac{4}{5}=-10\\ 2x=1-\left(-10\right)=11\\ x=\dfrac{11}{2}\\ g;\dfrac{1}{6}-\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{8}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{6}-\dfrac{1}{8}=\dfrac{1}{24}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{24}\Rightarrow x=\dfrac{3}{4}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{1}{24}\Rightarrow x=\dfrac{7}{12}\end{matrix}\right.\)