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\(x^2+6x+5=x^2+5x+x+5=x\left(x+5\right)+\left(x+5\right)=\left(x+1\right)\left(x+5\right)\)
\(x^2-7x+12=x^2-4x-3x+12=x\left(x-4\right)-3\left(x-4\right)=\left(x-3\right)\left(x-4\right)\)
\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
\(a,x^2+6x+5=x^2+5x+x+5\)
\(=x\left(x+5\right)+\left(x+5\right)=\left(x+5\right)\left(x+1\right)\)
\(b,\)\(x^2-7x+12=x^2-3x-4x+12\)
\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
\(c,\)\(x^2-7x+10=x^2-2x-5x+10\)
\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)

A = 1/(x - 2) + (x^2 - x - 2)/(x^2 - 7x + 10) - (2x - 4)/(x - 5)
A = 1/(x - 2) + ((x + 1)(x - 2))/((x - 2)(x + 5)) - (2x - 4)/(x - 5)
A = 1/(x - 2) + (x + 1)/(x - 5) - (2x - 4)/(x - 5)
A = 1/(x - 2) + (x - (2x - 4) + 1)/(x - 5)
A = 1/(x - 2) - 1

\(N=\dfrac{1}{x-2}+\dfrac{x^2-x-2}{x^2-7x+10}-\dfrac{2x-4}{x-5}\)
\(=\dfrac{x-5+x^2-x-2-2\left(x-2\right)^2}{\left(x-5\right)\left(x-2\right)}\)
\(=\dfrac{x^2-7-2\left(x^2-4x+4\right)}{\left(x-5\right)\left(x-2\right)}\)
\(=\dfrac{x^2-7-2x^2+8x-8}{\left(x-2\right)\left(x-5\right)}\)
\(=\dfrac{-x^2+8x-15}{\left(x-2\right)\left(x-5\right)}\)
\(=\dfrac{-\left(x-5\right)\left(x-3\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{3-x}{x-2}\)

a, A= -3x2 + 15x + 3x2 - 12x -3x + 10
A= 15x - 15x + 10
A= 10
b, B= 4x3 - 28x2 + 8x - 4x3 + 28x2 - 8x + 20
B= 20

\(C=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)
Đặt \(x^2-7x=t\),khi đó:
\(C=\left(t+10\right).\left(t-10\right)=t^2-10^2=t^2-100\)
Vì \(t^2\ge0=>t^2-100\ge-100\) (với mọi t)
Dấu "=" xảy ra\(< =>t=0< =>x^2-7x=0< =>x\left(x-7\right)=0< =>\orbr{\begin{cases}x=0\\x=7\end{cases}}\)
Vậy minC=-100 khi x=0 hoặc x=7
\(=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)
\(=\left(x^2-7x\right)^2-100\)