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\(A=\frac{1\cdot1}{1\cdot2}\cdot\frac{2\cdot2}{2\cdot3}\cdot\frac{3\cdot3}{3\cdot4}\cdot\frac{4\cdot4}{4\cdot5}=\frac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot3\cdot4}\cdot\frac{1\cdot2\cdot3\cdot4}{2\cdot3\cdot4\cdot5}=\frac{1}{5}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)
= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)\(+...+\frac{1}{2005}-\frac{1}{2006}\)
= \(\frac{1}{2}-\frac{1}{2006}\)
= \(\frac{501}{1003}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2005.2006}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(=\frac{1}{2}-\frac{1}{2006}\) >> Đúng 100% nha!! ^ ^
Bài 1
a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk mình nha!!
Câu 2:
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)
\(=\frac{3\cdot100}{2}\)
\(=\frac{300}{2}=150\)
Tìm \(x\) câu a:
\(\frac13.x\) + \(\frac25.\left(x+1\right)\) = 0
\(\frac{5}{15}x\) + \(\frac{6}{15}x\) + \(\frac25\) = 0
\(\frac{11}{15}x\) = - \(\frac25\)
\(x=-\frac25:\frac{11}{15}\)
\(x\) = - \(\frac25\times\frac{15}{11}\)
\(x\) = - \(\frac{6}{11}\)
Vậy \(x=-\frac{6}{11}\)
Tìm \(x\) câu b:
\(x\) x 25% = 0,5
\(x\times0,25\) = 0,5
\(x=0,5:0,25\)
\(x=2\)
Vậy \(x=2\)
Câu 1a:
1/3x + 2/5(x + 1) = 0
1/3x + 2/5x + 2/5 = 0
1/3x + 2/5x = - 2/5
x(1/3 + 2/5) = -2/5
x.(5/15 + 6/15) = -2/5
x.11/15 = - 2/5
x = - 2/5 : 11/15
x = - 6/11
Vậy x = -6/11
Câu b:
x . 25%. x = 0,5
x.x = 0,5 : 25%
x^2 = 2
x = - \(\sqrt2\); x = \(\sqrt2\)
Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )
Câu 1a:
1/3x + 2/5(x + 1) = 0
1/3x + 2/5x + 2/5 = 0
1/3x + 2/5x = - 2/5
x(1/3 + 2/5) = -2/5
x.(5/15 + 6/15) = -2/5
x.11/15 = - 2/5
x = - 2/5 : 11/15
x = - 6/11
Vậy x = -6/11
Câu b:
x . 25%. x = 0,5
x.x = 0,5 : 25%
x^2 = 2
x = - \(\sqrt2\); x = \(\sqrt2\)
Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )
\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+........+\frac{2}{x\left(x+1\right)}=\frac{2008}{2010}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+........+\frac{1}{x\left(x+1\right)}=\frac{2008}{4020}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{4020}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2008}{4020}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2008}{4020}\)
\(\frac{1}{x+1}=\frac{1}{2010}\)
=> x + 1 = 2010
=> x = 2010 - 1
=> x = 2009
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x\left(x+1\right)}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{2010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2010}\)
\(\Leftrightarrow x+1=2010\)
\(\Leftrightarrow x=2009\)