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\(\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{50}=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
=> \(2013x.\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2013x.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
=> \(2013x.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\Rightarrow2013x=2012\Rightarrow x=\frac{2012}{2013}\)
Vậy \(x=\frac{2012}{2013}\)
p/s: --trình bày sai sót mong bỏ qua
a) \(\dfrac{2}{3x+9}-\dfrac{x-3}{3x^2+9x}\)
\(=\dfrac{2}{3\left(x+3\right)}-\dfrac{x-3}{3x\left(x+3\right)}\)
\(=\dfrac{2x}{3x\left(x+3\right)}-\dfrac{x-3}{3x\left(x+3\right)}\)
\(=\dfrac{2x-x+3}{3x\left(x+3\right)}\)
\(=\dfrac{x+3}{3x\left(x+3\right)}\)
\(=\dfrac{1}{3x}\)
b) \(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}:\dfrac{3\left(x+1\right)}{5\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}:\dfrac{3\left(x+1\right)}{5\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}.\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\)
\(=\dfrac{x}{\left(x-1\right).3}\)
\(=\dfrac{x}{3x-3}\)
c) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+99}-\dfrac{1}{x+100}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+100}\)
\(=\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
\(=\dfrac{x+100-x}{x\left(x+100\right)}\)
\(=\dfrac{100}{x\left(x+100\right)}\)
\(A=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{2n+1}{n^2\cdot\left(n^2+2n+1\right)}\)
\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)
\(A=1-\dfrac{1}{n^2+2n+1}\)
\(A=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}...\dfrac{99}{100}.\dfrac{101}{100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
nói mk học lớp 7,ko bt lm bài của Nguyễn Tấn Dũng mà bài lớp 8 nào cũng làm đc.Bài toán khó nào cũng giải đc ,mà câu của Nguyễn Tấn Dũng thì bó tay ,thật ra cậu đang nói sạo hay thật z Nguyễn Huy Tú
???
ĐKXĐ: \(x\ne0;-1;-2;...;-100\)
Ta có: \(\dfrac{1}{x\left(x+1\right)}=\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{1}{x}-\dfrac{1}{x+1}\)
Tương tự, ta có biểu thức đã cho bằng:
\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+99}-\dfrac{1}{x+100}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+100}\)
a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)
\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)
b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)
\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)
Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)
c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)
Bài 58:
a, \(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\)
b, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\)
Vậy...
\(B=\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}\)
\(=\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}\)
\(=\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+7\right)+6\left(x+7\right)}\)
\(=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}\)
\(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}\)
\(=\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}\)
\(=\dfrac{3}{x^2+11x+28}\)
Vậy...
58,
\(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{x+1}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\)b,
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)\(=1-\dfrac{1}{n\left(n+1\right)}=\dfrac{n^2+n-1}{n\left(n+1\right)}\)
\(B=\dfrac{1}{\left(x^2+9x+20\right)}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}\)\(=\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\dfrac{1}{\left(x^2+6x\right)+\left(7x+42\right)}\)\(=\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+6\right)+7\left(x+6\right)}\)\(=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}\)\(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x-7}\)\(=\dfrac{1}{x+4}-\dfrac{1}{x+7}\)