\(3x^2-5x+2\)

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

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

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

Đề sai rồi bạn phải + 2 chứ

Bạn đưa ra 1 ví dụ đi rồi mình giảng

\(3x^3-7x^2+17x-5\)

\(=3x^3-6x^2-x^2+15x+2x-5\)

\(=\left(3x^3-6x^2+15x\right)-\left(x^2-2x+5\right)\)

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

\(=\left(3x-1\right)\left(x^2-2x+5\right)\)

có cách nào khác ngoài phương pháp tách không bạn

3x³-7x²+17x-5

=3x³-x²-6x²+2x+15x-5

=x².(3x-1)-2x.(3x-1)+5.(3x-1)

=(3x-1)(x²-2x+5)

Giải thích: `x^2-5x+1`

`=x^2-2. 5/2x+25/4-21/4`

`=(x-5/2)^2-21/4`

`=(x-5/2-\sqrt{21}/2)(x-5/2+\sqrt{21}/2)`

`=(x-[5+\sqrt{21}]/2)(x-[5-\sqrt{21}]/2)`

\(H=3x-5-3x^2\)

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

\(=-3\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{17}{12}\right)\)

\(=-3\left(\left(x-\frac{1}{2}\right)^2+\frac{17}{12}\right)\)

\(=\frac{-17}{4}-3\left(x-\frac{1}{2}\right)^2\le\frac{-17}{4}\)

\(MAXH=\frac{-17}{4}\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

\(8x^3+12x^2y+6xy^2+y^3-z^3\)

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

\(=\left(2x+y-z\right)\left[4x^2+z\left(2x+y\right)+z^2\right]\)

a, 8a3 - 36a2 +54ab2 - 27b3

=(8a3-36a2b +54ab2 - 27b3)

=(2a-3b)2

=(2a-3b)(2a-3b)(2a-3b)

b, 8x3 + 12x2y + 6xy2 + y3 - z 3

=(8x3 + 12x2y + 6xy2 + y3) - z3

=(2x + y)3 - y3

=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2

= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2

\(xy^2-5xz=x\left(y^2-5z\right)\)

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

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