Question

Viết các đa thức sau thành tích:

\(16-x^2\)

\(4x^2-9\)

\(a^4-25\)

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

\(\left(a+b\right)^2-\left(m-n\right)^2\)

\(x^3-27\)

\(64x^3+\frac{1}{27}\)

Answer 1

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

\(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2-5\right)\left(a^2+5\right)\)

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

\(\left(a+b\right)^2-\left(m-n\right)^2=\left(a+b-m+n\right)\left(a+b+m-n\right)\)

\(x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+3^2\right)\)

\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2+\frac{4}{3}x+\frac{1}{9}\right)\)

Tham khảo~

Answer 2

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

\(4x^2-9=\left(2x\right)^2-3^2=\left(2x+3\right)\left(2x-3\right)\)

\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2+5\right)\left(a^2-5\right)\)

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

\(\left(a+b\right)^2-\left(m-n\right)^2=\left(a+b+m-n\right)\left(a+b-m+n\right)\)

\(x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+9\right)\)

\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2-\frac{4}{3}x+\frac{1}{9}\right)\)