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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: \(=\dfrac{4}{x+2}-\dfrac{3}{x-2}+\dfrac{12}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
b: \(=\dfrac{6x+3\left(x-1\right)+2\left(x-2\right)}{6}=\dfrac{6x+3x-3+2x-4}{6}=\dfrac{11x-7}{6}\)
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
a: \(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot...\cdot\left(3^{1024}+1\right)}{8}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\cdot...\cdot\left(3^{1024}+1\right)}{8}\)
\(=\dfrac{3^{2048}-1}{8}\)
b: \(=100+99+98+97+...+50+49\)
Số số hạng là (100-49):1+1=100-48=52 số
Tổng là (100+49)*52/2=149*26=3874
c: \(=x^2-2x+1+x^2-4-x^3-9x^2-27x-27\)
\(=-x^3-7x^2-29x-30\)
a) \(\frac{3x^2+5x+1}{x^3-1}-\frac{1-x}{x^2+x+x}-\frac{3}{x-1}\)
\(=\frac{3x^2+5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(1-x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{3x^2+5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{1-x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2+3x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{3x^2+5x+1-1+x^2-3x^2-3x-3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^2+2x-3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^2+3x-x-3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x+3}{x^2+x+1}\)
công thức tổng quát (n+1-n)/n(n+1)
a.)1/x(x+1)+1(x+1)(x+2)+...+1/(x+99)(x+110
=1/x-1(x+1)+1/(x+1)-1/(x+1)+...+1/(x+99)-1/(x+100)
=1/x-1/(x+100)
=(x+100-x)/x(x+100)
=100/x(x+100)
b;)1/(x-1)(x-2)+2/(x-2)(x-3)-3/(x-3)(x-1)
=(x-3)/(x-1)(x-2)(x-3)+(2x-1)/(x-1)(x-2)(x-3)-(3x-6)/(x-1)(x-2)(x-3)
=(x-3+2x-1-3x+6)/(x-1)(x-2)(x-3)
=2/(x-1)(x-2)(x-3)
Câu a không hiểu cho lắm