Tính
(x-3)+(x-2)+(x-1)+...+(x+10)
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\(\dfrac{1}{x+1}:\dfrac{x+2}{x+1}:\dfrac{x+3}{x+2}:...:\dfrac{x+10}{x+9}\)
\(=\dfrac{1}{x+1}\cdot\dfrac{x+1}{x+2}\cdot\dfrac{x+2}{x+3}\cdot...\cdot\dfrac{x+9}{x+10}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\cdot...\cdot\left(x+9\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\cdot...\cdot\left(x+9\right)\left(x+10\right)}\)
\(=\dfrac{1}{x+10}\)
1) 9 x 10 + 10 x 11 + 11 x 12 + ....+ 2015 x 2016
=9x10x11-8x9x10+10x11x12-9x10x11+...+2015x2016x2017-2014x2015x2013
=2015x2016x2017-8x9x10
=8193537360
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\dfrac{1}{2}-\dfrac{1}{50}=\dfrac{12}{25}\)
a: =>(x+10)(x-1)=0
=>x=-10 hoặc x=1
b: \(A=x^3-1-\left(x+5\right)\left(x^2-3\right)-5x^2-10x-5\)
\(=x^3-5x^2-10x-6-x^3+3x-5x^2+15\)
=-7x+9
=110/13
\(S=\frac{1}{x\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)
\(=\frac{1}{2}\left[\frac{2}{x\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{2}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\right]\)
\(=\frac{1}{2}\left[\frac{x+2-x}{x\left(x+1\right)\left(x+2\right)}+\frac{\left(x+3\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{\left(x+10\right)-\left(x+8\right)}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\right]\)
Có: (x-3)+(x-2)+(x-1)+...+(x+10)
=> x-3 +x-2 +x-1 +...+x+9+x+10
=> x +x +x +x+...+x +(-3+-2+-1+0+...+9+10)
Số số hạng x là: [10-(-3):1+1]=14(số)
=> 14x + [(10+-3):2 x 14]= 14x + 49
=7.2.x+7.7= 7(2x+7)