Question

Phân tích đa thức thành nhân tử 

4x²-12x+5

Tìm x biết (x+1)(x+2)-(x+2)(x+3)=0

Answer 1

\(\left(x+1\right)\left(x+2\right)-\left(x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+1-x-3\right)=0\)

\(\Leftrightarrow-2\left(x+2\right)=0\)

\(\Leftrightarrow x=-2\)

Answer 2

\(4x^2-12x+5=4x^2-10x-2x+5=2x\left(2x-5\right)-\left(2x-5\right)=\left(2x-1\right)\left(2x-5\right)\)

\(\left(x+1\right)\left(x+2\right)-\left(x+2\right)\left(x+3\right)=0\)

\(\left(x+2\right)\left(-2\right)=0\)\(\Rightarrow x+2=0\) hay \(x=-2\)

Answer 3

\(4x^2-12x+5=4\left(x^2-3x+\frac{5}{4}\right)=4\left(x^2-3x+\frac{9}{4}-\frac{9}{4}+\frac{5}{4}\right)=4\left(x-\frac{3}{2}\right)^2-2=\left(2x-1\right)\left(2x-5\right)\)

Answer 4

cảm mơn mb nhìu lắm 

Answer 5

\(4x^2-12x+5=\left(2x\right)^2-2.2x.3+9-4\)

                            \(=\left(2x\right)^2-2.2x.3+3^2-2^2\)

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

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

                             \(=\left(2x-5\right)\left(2x-1\right)\)