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26 tháng 12 2016

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)

26 tháng 12 2016

Câu a không hiểu cho lắm

5 tháng 10 2021

Bài 2:

a) \(=x^2-36y^2\)

b) \(=x^3-8\)

Bài 3:

a) \(=x^2+2x+1-x^2+2x-1-3x^2+3=-3x^2+4x+3\)

b) \(=6\left(x-1\right)\left(x+1\right)=6x^2-6\)

a: \(=\dfrac{x^2-2x+1}{x}:\dfrac{x-1-3x^2+3x-3}{\left(x-1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{-2x^2+4x-4}\)

\(=\dfrac{\left(x-1\right)^3\cdot\left(x^2-x+1\right)}{-2x\left(x^2-2x+2\right)}\)

b: \(=\left[\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right]:\dfrac{2}{x^2+1}\)

\(=\dfrac{x^3-3x^2+3x+1+2x^2-4x+1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)

\(=\dfrac{x^3+3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)

23 tháng 9 2023

\(\dfrac{2x+4}{x^3-1}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\\ =\dfrac{2x+4}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\\ =\dfrac{2x+4}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{2x+4-2x^2-2x-2+x^2-x+2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{-x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{-x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=-\dfrac{x}{x^2+x+1}\)

23 tháng 9 2023

`a, 2/(x+1)` hay `2/(x-1)` cậu nhỉ?

`b,`

\(\dfrac{x-1}{x^2-5x+6}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(x-2\right)^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x-1-\left(x^2-6x+9\right)+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{x-1-x^2+6x-9+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{3x-6}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{3}{x-3}\)

26 tháng 12 2021

a)\(\dfrac{x^2}{x-1}+\dfrac{1-2x}{x-1}\)

=\(\dfrac{x^2+1-2x}{x-1}\)

=\(\dfrac{x^2-2x+1}{x-1}\)

=\(\dfrac{\left(x-1\right)^2}{x-1}\)

= x - 1

 

26 tháng 12 2021

b) \(\dfrac{x}{x-3}\) + \(\dfrac{-9}{x^2-3x}\)

=\(\dfrac{x}{x-3}\)\(\dfrac{-9}{x\left(x-3\right)}\)

=\(\dfrac{x.x}{x\left(x-3\right)}\) + \(\dfrac{-9}{x\left(x-3\right)}\)

=\(\dfrac{x^2+3^2}{x\left(x-3\right)}\)

=\(\dfrac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}\)

=\(\dfrac{x+3}{x}\)

#Fiona

 

23 tháng 7 2023

\(a,=\left(\dfrac{1-x}{x}+\dfrac{x^3-x}{x}\right)\times\dfrac{x}{x-1}\\ =\dfrac{1-x+x^3-x}{x}\times\dfrac{x}{x-1}\\ =\dfrac{1-2x+x^3}{x-1}\\ b,=\left(\dfrac{x-x^2}{x.x^2}\right).\dfrac{x^2}{y}+\dfrac{x}{y}\\ =\dfrac{x-x^2}{xy}+\dfrac{x}{y}\\ =\dfrac{x-x^2+x^2}{xy}=\dfrac{x}{xy}=\dfrac{1}{y}\)

\(c,=\dfrac{3}{x}-\dfrac{2}{x}\times x+\dfrac{x}{3}\\ =\dfrac{3}{x}-2+\dfrac{x}{3}\\ =\dfrac{3-2x+x^2}{3x}\)

\(\left(\dfrac{1}{x}+x-2\right):\left(\dfrac{1}{x^2-x}+1-\dfrac{3}{x-1}\right)\)

\(=\dfrac{x^2-2x+1}{x}:\dfrac{1+x^2-x-3x}{x\left(x-1\right)}\)

\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{x\left(x-1\right)}{x^2-4x+1}=\dfrac{\left(x-1\right)^3}{x^2-4x+1}\)

24 tháng 12 2021

b: \(=\dfrac{7x-42-x^2+36}{x\left(x-6\right)}=\dfrac{-x^2+7x-6}{x\left(x-6\right)}=\dfrac{-x+1}{x}\)

24 tháng 12 2021

\(\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x-3\right)}=\dfrac{x\left(x+3\right)-3\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+3x-3x-9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x}\)

a: \(\left(x^2-2x+2\right)\left(x^2+2x+2\right)\left(x-2\right)\left(x+2\right)\)

\(=\left[\left(x^2+2\right)^2-4x^2\right]\left(x^2-4\right)\)

\(=\left(x^3+2x\right)^2-4\left(x^2+2\right)^2-4x^4+16x^2\)

\(=x^6+4x^4+4x^2-4x^4+16x^2-4\left(x^4+4x^2+4\right)\)

\(=x^6+20x^2-4x^4-16x^2-16\)

\(=x^6-4x^4+4x^2-16\)

b: \(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)

\(=3x^3+6x-3x^3+3x\)

=9x

d: \(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=100+99+...+2+1\)

=5050