Question

Sử dụng hằng đẳng thức , phân tích đa thức thành nhân tử 

\(\frac{1}{64}x^6-125y^3\)

\(27a^3-54a^2b+36ab^2-8b^3\)

\(x^6-x^6\)

\(10x-25-x^2\)

Answer 1

\(a,\frac{1}{64}x^6-125y^3\)

\(=\left(\frac{1}{2}x\right)^6-\left(5y\right)^3\)

\(=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)

\(\left(\frac{1}{4}x^2-5y\right)\left[\left(\frac{1}{4}x^2\right)^2+\left(\frac{1}{4}x^2\right).5y+25y^2\right]\)

\(b,27a^3-54a^2b+36ab^2-8b^3\)

\(=\left(3a\right)^3-3.2.\left(3a\right)^2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)

\(=\left(3a-2b\right)^3\)

\(c,x^6-x^6\)

\(=0\)

\(d,10x-25-x^2\)

\(=-x^2+10x-25\)

\(=-\left(x^2-10x+25\right)\)

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