a) x(x + 4)(x + 6)(x + 10) + 128 = [x(x + 10)][(x + 4)(x + 6)] + 128
\(=\left(x^2+10x\right)+\left(x^2+10x+24\right)+128\)
Đặt x2 +10x + 12 = y , đa thức có dạng :
\(\left(y-12\right)\left(y+12\right)+128=y^2-144+128=y^2-16=\left(y+4\right)\left(y-4\right)\)
\(=\left(x^2+10x+8\right)\left(x^2+10x+16\right)=\left(x+2\right)\left(x+8\right)\left(x^2+10x+8\right)\)
b) Giả sử x \(\ne\) 0 , ta viết :
Đặt x - \(\frac{1}{x}\) = y thì
\(x^2+\frac{1}{x^2}=y^2+2\) , do đó :
\(A=x^2\left(y^2+2+6y+7\right)=x^2\left(y+3\right)^2=\left(xy+3x\right)^2=\left[x\left(x-\frac{1}{x}\right)^2+3x\right]^2=\left(x^2+3x-1\right)^2\)
c) \(\left(a+b+c\right)^3-4\left(a^3+b^3+c^3\right)-12abc\)
Đặt a + b = m , a - b = n thì 4ab = m2 - n2
\(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=m\left(n^2+\frac{m^2-n^2}{4}\right)\) . Ta có :
\(C=\left(m+c\right)^3-4.\frac{m^3+3mn^2}{4}-4c^3-3c\left(m^2-n^2\right)=3\left(-c^3+mc^2-mn^2+cn^2\right)\)
\(=3\left[c^2\left(m-c\right)-n^2\left(m-c\right)\right]=3\left(m-c\right)\left(c-n\right)\left(c+n\right)=3\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)
