cho đề hay cho phương pháp giải?

1 cách thoi:

x²+5x+4

= x²+x+4x+4

= x(x+1)+4(x+1)

= (x+1)(x+4)

x²+5x+6

= x²+2x+3x+6

= x(x+2)+3(x+2)

= (x+2)(x+3)

(a+b)³+(a-b)³=(a³+3a²b+3ab²+b³)+(a³-3a²b+3ab²-b³)

                   =a⁶+6a²b⁴

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

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

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

\(=a^2b^2+1+a^2+b^2\)

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

\(x^{n+3}+x^n=x^n.x^3+x^n=x^n\left(x^3+1\right)=x^n\left(x+1\right)\left(x^2-x+1\right)\)

\(x^{n+3}+x^n=x^n\left(x^3+1\right)=x^n\left(x+1\right)\left(x^2-x+1\right)\)

\(x^{n+3}+x^n=x^n\left(x^3+1\right)=x^n\cdot\left(x+1\right)\left(x^2-x+1\right)\)

\(x^3-2x^2-19x+20\)

\(=x^3+3x^2-4x-5x^2-15x+20\)

\(=\left(x^3+3x^2-4x\right)-\left(5x^2+15x-20\right)\)

\(=x\left(x^2+3x-4\right)-5\left(x^2+3x-4\right)\)

\(=\left(x^2+3x-4\right)\left(x-5\right)\)

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

\(=\left[x\left(x+4\right)-\left(x+4\right)\right]\left(x-5\right)\)

\(=\left(x+4\right)\left(x-5\right)\left(x-1\right)\)

x^3-2x^2-19x+20

=x^3-5x^2+3x^2-15x-4x+20

=(x-5)(x^2+3x-4)

=(x+4)(x-1)(x-5)

Có: x³ - 2x² - 19x + 20

=> x³ - 2x² + x - 20x + 20

=> x(x² - 2x + 1) - 20(x - 1)

=> x(x - 1)² - 20(x - 1)

=> (x - 1)[x(x - 1) - 20]

=> (x - 1)(x² - x - 20)

=> (x - 1)(x² - 5x + 4x - 20)

=> (x - 1)[x(x - 5) + 4(x - 5)]

=> (x - 1)(x + 4)(x - 5)