Chia nhỏ ra cậu ơi :v

Cậu đặt câu hỏi free nên đặt nhỏ ra thì mới có người làm nha để như này dày cộp không ai dám làm đou =(((

cảm ơn nhé

a) (x - 2)(x - 3).                        b) 3(x - 2)(x + 5).

c) (x - 2)(3x + 1).                     d) (x-2y)(x - 5y).

e) (x + l)(x + 2)(x - 3).             g) (x-1)(x + 3)( x 2  + 3).

h) (x + y - 3)(x - y + 1).

a, \(x^2-5x+6=x^2+x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)

b, \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)

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

c, \(x^2+7x+10=x^2+2x+5x+10=x\left(x+2\right)+5\left(x+2\right)=\left(x+2\right)\left(x+5\right)\)

a) x² - 5x + 6 = (x²-2x)-(3x-6)=x(x-2)-3(x-2)=(x-3)(x-2)

b) 3x² + 9x -30= 3(x²+3x-10) = 3((x²+5x)-(2x+10)) = 3(x(x+5)-2(x+5)) = 3(x-2)(x+5)

c) x² + 7x + 10 =( x²+5x)+(2x+10)=x(x+5)+2(x+5)=(x+2)(x+5)

bạn tách ra từng ít câu 1 thôi ạ

a: \(=5x\left(xy^2+3x+6y^2\right)\)

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

c: \(=\left(x-3\right)\left(x-4\right)\)

d: \(=x\left(x^2-2xy+y^2-9\right)\)

=x(x-y-3)(x-y+3)

e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)

f: \(=\left(x-4\right)\left(x+3\right)\)

a) (x - 1)(x - 2).                        b) 4(x - 2)(x - 7).

c) (x + 2)(2x +1).                    d) (x - l)(2x - 7).

e) (2x + 3y - 3)(2x - 3y +1).    g) (x - 3)( x 3   +   x 2  - x +1).

h) (x + y)(x + y-l)(x + y + l).

Lời giải của các bạn đều thỏa mãn yêu cầu đề bài là phân tích đa thức thành nhân tử

\(1,\\ 12x^6y^3:4x^3y=3x^3y^2\\ \left(x+1\right)\left(x^2-x+1\right)=x^3+1\\ 2x^2y\left(x^2+3xy\right)=3x^4y+6x^3y^2\\ 2,\\ a,=2xy\left(2x+3y-4\right)\\ b,=\left(x-3\right)\left(x+y\right)\\ c,=\left(x-2\right)\left(x+2\right)+y\left(x-2\right)=\left(x+y+2\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ 3,\\ a,\Leftrightarrow x^2-x^2+2x=2\\ \Leftrightarrow2x=2\Leftrightarrow x=1\\ b,\Leftrightarrow\left(x-2\right)\left(x-2+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a: \(x^2-6x+5=\left(x-5\right)\left(x-1\right)\)

b: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

c: \(x^2+8x+15=\left(x+5\right)\left(x+3\right)\)

d: \(2x^2-5x-12=\left(x-4\right)\left(2x+3\right)\)

e: \(x^2-13x+36=\left(x-9\right)\left(x-4\right)\)

\(a,=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ b,=4x^2\left(x^2+2x+1\right)=4x^2\left(x+1\right)^2\\ c,=xy^2\left(x^2-2xy+y^2\right)=xy^2\left(x-y\right)^2\\ d,=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\\ e,=\left(5x-2y\right)\left(5x+2y\right)\\ f,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ i,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)

-Đặt \(t=\left(x^2-x+1\right)\)

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

\(=t^2-5xt+4x^2\)

\(=t^2-4xt-xt+4x^2\)

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

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

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

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

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

CAM ON - HOANG