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a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
1/1+(-2)+3+(-4)+.....+19+(-20)
=1-2+3-4+.....+19-20
=(1+3+.....+19)-(2+4+.....+20)
={(19+1).[(19-1):2+1]:2}-{(20+2).[(20-2):2+1]:2}
={20.10:2}-{22.10:2}
=10:2.(20-22)
=5.(-2)
=-10
\(\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+...+\frac{1}{9}.\frac{1}{10}\)
\(=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{10}\)
\(=\frac{9}{10}\)
Ta có :
S = 1/1+2 + 1/1+2+3 +...+1/1+2+3+...+9+10
S = 1/3 + 1/6 + 1/10 + ... + 1/45
S = 1/1.3 + 1/2.3 + ... + 1/5.9
S:2 = 1/1.2.3 + 1/2.2.3 + ... + 1/2.5.9
S:2 = 1/2.3 + 1/3.4 + ... + 1/9.10
S:2 = 3-2/2.3 + 4-3/3.4 + ... + 10-9/9.10
S:2 = 3/2.3 - 2/2.3 + ... + 10/9.10 - 9/9.10
S:2 = 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10
S:2 = 1/2 - 1/10
S:2 = 4/10
=> S = 4/5
(Chó muốn sủa thì cứ sủa đi , có ai cấm chó sủa đâu ^.^)