Question

1.Phân tích thành nhân tử (phương pháp tách 1 hạng tử thành nhiều hạng tử)

a. x^2 - 9x + 20

b .x^2 + 9x + 20

c. x^2 + x - 20

d. x^2 - x - 20

Answer 1

\(\text{a) }x^2-9x+20\)

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

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

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

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

\(\text{b) }x^2+9x+20\)

\(=x^2+4x+5x+20\)

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

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

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

\(\text{c) }x^2+x-20\)

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

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

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

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

\(\text{d) }x^2-x-20\)

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

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

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

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