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c: \(=\dfrac{8}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x^2+3\right)\left(x-1\right)}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x-1}\)
x4+x2+1
=(x2)2+2x2+1-2x2+x2
=(x2+1)2-2x2+x2
= (x² + 1)² − x²
= (x² + x+ 1 )(x² − x+ 1 )
\(x^4+x^2+1\)
\(=\left[\left(x^2\right)^2+2.x^2.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]-\left(\frac{1}{2}\right)^2+1\)
\(=\left(x^2+\frac{1}{2}\right)^2-\frac{1}{4}+\frac{4}{4}\)
\(=\left(x^2+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(=\dfrac{x^2+2+5x-10}{x^2-4}=\dfrac{x^2+5x-8}{\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{x}{x-y}-\dfrac{1}{x-y}-\dfrac{1-y}{y-x}=\dfrac{x}{x-y}-\dfrac{1}{x-y}+\dfrac{y-1}{x-y}=\dfrac{x-1+y-1}{x-y}=\dfrac{x+y-2}{x-y}\)
a) 16x2-(x2+4)2= (4x)2-(x2+4)2
= (4x-x2-4)(4x+x2+4)
\(\text{b) 27x^3-54x^2+36x-8=[(3x)^3-3.(3x)^2.2+3.3x.2^2-2^3}]\)
= (3x-2)3
\(\text{c) (x+y)^3 - (x-y)^3= (x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]}\)
=2y(x2+2xy+y2+x2-y2+x2-2xy+y2)
= 2y(3x2+y2)
1. = (x-2)^2 - y^2 = (x - 2 - y)(x-2+y)
2. = (x-y-x-y)(x-y+x+y) = 2(-y)2x = -4xy
x+\(\dfrac{y}{2}\)+x+\(\dfrac{2}{2}\)x2+4
=2x+\(\dfrac{4+y}{2}\)+4