x2+5x-6
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= ab^2 - ac^2 + bc^2 - a^2b + a^2c -cb^2
=ab(b-a) - ac(c-a) - bc(b-c)
=ab(b-a) - ac(c-a) - bc(b-c)
=ab(b-a) - ac(-a+b-b+c) - bc(b-c)
=ab(b-a) - ac(b-a) + ac(b-c) - bc(b-c)
=(b-a)(ab-ac) - (b-c)(ac-bc)
=a(b-c)(b-a) - ac(a-b)(b-c)
=(b-c)[a(b-a) - c(b-a)]
=(b-c)(b-a)(a-c)
a, \(\frac{3x}{2x+4};\frac{x+3}{x^2-4}\)
Ta có : \(2x+4=2\left(x+2\right)\)
\(x^2-4=\left(x-2\right)\left(x+2\right)\)
MTC : \(2\left(x-2\right)\left(x+2\right)\)
\(\frac{3x}{2x+4}=\frac{3x}{2\left(x+2\right)}=\frac{3x\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}=\frac{3x^2-6x}{2\left(x-2\right)\left(x+2\right)}\)
\(\frac{x+3}{x^2-4}=\frac{x+3}{\left(x-2\right)\left(x+2\right)}=\frac{2x+6}{\left(x-2\right)\left(x+2\right)}\)
c, \(\frac{2x}{x^2-8x+16};\frac{x}{3x^2-12x}\)
Ta có : \(x^2-8x+16=\left(x-4\right)^2\)
\(3x^2-12x=3x\left(x-4\right)\)
MTC : \(3x\left(x-4\right)^2\)
\(\frac{2x}{x^2-8x+16}=\frac{2x}{\left(x-4\right)^2}=\frac{6x^2}{3x\left(x-4\right)^2}\)
\(\frac{x}{3x^2-12x}=\frac{x}{3x\left(x-4\right)}=\frac{x^2+4x}{3x\left(x-4\right)\left(x+4\right)}\)