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a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)
b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
a, -5/6 -x = 7/12 + -1/3
⇔-10/12 - 12x/12 = 7/12 + -4/12
⇒-10 - 12x = 7 - 4
⇔-12x = 7 - 4 +10
⇔-12x = 13
⇔x = -13/12
b, x+13/-15 = 1/3
⇔-(x+13)/15 = 5/15
⇒ -x - 13 = 5
⇔-x = 5 +13
⇔-x = 18
⇔x = -18
c,-15/x-1 = -3/5
⇔-75/(x-1).5 = -3.(x-1)/5.(x-1)
⇒-75 = -3x + 3
⇔3x = 3 + 75
⇔3x = 78
⇔x = 26
d, (1/2).x + -2/5 = 1/5
⇔5x/10 + -4/10 = 1/10
⇒5x - 4 = 1
⇔5x = 1 + 4
⇔5x = 5
⇔x = 1
e, (-2/3).x + 1/5 = 1/10
⇔-20x/30 + 6/30 = 3/30
⇒-20x + 6 = 3
⇔-20x = 3 - 6
⇔-20x = -3
⇔x = 3/20
f, 4/5 - (1/2).x = 1/10
⇔8/10 - 5x/10 = 1/10
⇒8 - 5x = 1
⇔-5x = 1 - 8
⇔-5x = -7
⇔x=7/5