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,
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)\(=1-\dfrac{1}{n\left(n+1\right)}=\dfrac{n^2+n-1}{n\left(n+1\right)}\)
\(B=\dfrac{1}{\left(x^2+9x+20\right)}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}\)\(=\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\dfrac{1}{\left(x^2+6x\right)+\left(7x+42\right)}\)\(=\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+6\right)+7\left(x+6\right)}\)\(=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}\)\(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x-7}\)\(=\dfrac{1}{x+4}-\dfrac{1}{x+7}\)
