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\(\frac{x^4-y^4}{y^3-x^3}=\frac{\left(x^2\right)^2-\left(y^2\right)^2}{\left(y-x\right)\left(y^2+xy+x^2\right)}=-\frac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}=-\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=-\frac{\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
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\(=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}=\dfrac{2a\left(a-b\right)}{\left(a-b\right)\left(c+d\right)}=\dfrac{2a}{c+d}\)
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ĐKXĐ: \(x\ne2\)
\(\frac{3x-6}{x-2}\)
\(=\frac{3.\left(x-2\right)}{x-2}\)
\(=3\)
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Câu a :
\(\dfrac{3x^2-12x+12}{x^4-8x}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
Câu b :
\(\dfrac{7x^2+14x+7}{3x^2+3x}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
\(\frac{a^3+1}{a^2+3a+4}=\frac{\left(a+1\right)\left(a^2-a+1\right)}{a^2+3a+4}=\left(a^2-a+1\right)\left(\frac{a+1}{a^2+3a+4}\right)\)