a) \(4x^3\left(x^2+x\right)-\left(x^2+x\right)=\left(x^2+x\right)\left(4x^3-1\right)\)

b)\(\left(1-2a+a^2\right)-\left(b^2-2bc+c^2\right)=\left(1-a\right)^2-\left(b-c\right)^2=\)\(\left(1-a+b-c\right)\left(1-a-b+c\right)\)

lm tiếp câu c

c)  \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)

\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)

\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)

Đặt   \(x^2-9x+17=a\) ta có:

        \(C=\left(a-3\right)\left(a+3\right)-72\)

            \(=a^2-9-72\)

           \(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được:  \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)

\(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x+4\right)^2-\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10+16+8\right)+16\)

\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)

\(=\left(x^2+10x+16+4\right)^2\)

\(=\left(x^2+10+20\right)^2\)

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right) \left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\)
\(\left(1\right)\Rightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\)
\(=\left(x^2+10x+20\right)^2\)
 

\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+16\)

\(=\left[\left(x-2\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-6\right)\right]+16\)

\(=\left(x^2-10x+16\right)\left(x^2-10x+24\right)+16\)(1) 

Đặt \(x^2-10x+20=t\)thay vào (1) ta được : 

\(\left(t-4\right)\left(t+4\right)+16\)

\(=t^2-16+16\)

\(=t^2\)Thay \(t=x^2-10x+20\)ta được :

\(\left(x^2-10x+20\right)^2\)

\(=\left(x^2-2.5.x+25-25+20\right)^2\)

\(=\left[\left(x-5\right)^2-5\right]^2\)

\(=\left(x-5-\sqrt{5}\right)^2\left(x-5+\sqrt{5}\right)^2\)