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=(a+b+c)(a2+b2+c2−ab−bc−ca)
=(a+b+c)(a2+2ab+b2−ab−ac+c2)−3ab(a+b+c)
=(a+b)3+c3−3ab(a+b+c)
=a3+3ab(a+b)+b3+c3−3abc−3ab(a+b
a3+b3+c3−3abc
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(x - 1/2 )(x + 1/2 )(4x - 1)
= ( x 2 + 1/2 x - 1/2 x - 1/4 )(4x - 1)
= ( x 2 - 1/4 )(4x - 1)
= 4 x 3 – x 2 – x + 1/4
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\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right).....\left(1-\dfrac{1}{2008^2}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{2008^2-1}{2008^2}\)
\(=\dfrac{1.3}{4}.\dfrac{2.4}{9}.\dfrac{3.5}{16}....\dfrac{2007.2009}{2008^2}\)
\(=\left(\dfrac{1.2.3...2007}{2.3.4....2008}\right).\dfrac{3.4.5...2009}{2.3.4...2008}\)
\(=\dfrac{1}{2008}.\dfrac{2009}{2}=\dfrac{2009}{4016}\)
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\(=\dfrac{x^2}{x-1}-\dfrac{2x-1}{x-1}=\dfrac{x^2-2x+1}{x-1}=x-1\)
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\(\left(\dfrac{1}{x}+x-2\right):\left(\dfrac{1}{x^2-x}+1-\dfrac{3}{x-1}\right)\)
\(=\dfrac{x^2-2x+1}{x}:\dfrac{1+x^2-x-3x}{x\left(x-1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{x\left(x-1\right)}{x^2-4x+1}=\dfrac{\left(x-1\right)^3}{x^2-4x+1}\)
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Đặt \(x^2+1=a\)
Ta có: \(\dfrac{1}{x^2-x+1}-\dfrac{x^2+2}{x^2+1}+1\)
\(=\dfrac{1}{a-x}+\dfrac{a+1}{a}+1\)
\(=\dfrac{a}{a\left(a-x\right)}+\dfrac{\left(a+1\right)\left(a-x\right)}{a\left(a-x\right)}+\dfrac{a\left(a-x\right)}{a\left(a-x\right)}\)
\(=\dfrac{a+a^2-ax+a-x+a^2-ax}{a\left(a-x\right)}\)
\(=\dfrac{2a^2+2a-2ax-x}{a\left(a-x\right)}\)
\(=\dfrac{2\left(x^2+1\right)^2+2\left(x^2+1\right)-2x\left(x^2+1\right)-x}{\left(x^2+1\right)\left(x^2+1-x\right)}\)
\(=\dfrac{2\left(x^4+2x^2+1\right)+2x^2+2-2x^3-2x-x}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{2x^4+4x^2+2+2x^2+2-2x^3-3x}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{2x^4-2x^3+6x^2-3x+4}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
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(x – 1)(x + 1)(x + 2)
= ( x 2 + x – x – 1)(x + 2)
= ( x 2 – 1)(x + 2)
= x 2 ( x + 2) – 1.(x +2)
= x 3 + 2 x 2 – x – 2
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(1/2 x – 1) (2x – 3)
= x 2 - 3/2 x – 2x + 3
= x 2 - 7/2 x + 3