Question

[pham thuy duong]

phân tích đa thức thanahf nhân tử

\a, 4x^2-4x+4=0

\b, x^3-27

\c, x^3-4x^2+3x

\d,4x^2-12x+3

Answer 1

[tớ đây giốt lắm]

a)\(4x^2-4x+4=0\Leftrightarrow\left(2x-1\right)^2+3\) (đến đây hết pt dc rùi)

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

c)\(x^3-4x^2+3x=x^3-x^2-3x^2+3x\)

                                   =\(x^2\left(x-1\right)-3x\left(x-1\right)\)

                                 =\(x\left(x-3\right)\left(x-1\right)\)

d)\(4x^2-12x+3=\left(2x-3\right)^2-6\)

                                    =\(\left(2x-3\right)^2-\sqrt{6^2}\)

                                 =\(\left(2x-3-\sqrt{6}\right)\left(2x-3+\sqrt{6}\right)\)

Answer 2

[Huyền Nhi]

\(a,4x^2-4x+4=4\left(x^2-x+1\right)\)

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

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

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

                             \(=x\left[x\left(x-1\right)-3\left(x-1\right)\right]\)

                             \(=x\left(x-1\right)\left(x-3\right)\)

\(d,4x^2-12x+3=4\left(x^2-3x+\frac{3}{4}\right)\)

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

\(=4\left[\left(x-\frac{3}{2}\right)^2-\frac{3}{2}\right]\)

\(=4\left[\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^2\right]\)

\(=4\left(x-\frac{3}{2}-\frac{\sqrt{3}}{\sqrt{2}}\right)\left(x-\frac{3}{2}+\frac{\sqrt{3}}{\sqrt{2}}\right)\)

\(=4\left(x-\frac{3+\sqrt{6}}{2}\right)\left(x-\frac{3-\sqrt{6}}{2}\right)\)

P/s: Dương: câu d t k chắc nx, sai thì thông cảm :)) -Huyền Nhi-