\(15x^3y^2+10x^2y^2-2x^2y^3\)

\(=x^2y^2\left(15x+10-2y\right)\)

`15x^3 y^2 + 10x^2 y^2 - 2x^2 y^3`

`=x^2y^2(15x+10-2y)`

a: \(=5xy\left(x^2-2xy+y^2\right)=5xy\left(x-y\right)^2\)

b: \(=2x^2+10x-3x-15\)

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

1) \(3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)\)

2) \(5x^2-5y^2-10x+10y=5\left(x^2-y^2\right)-10\left(x-y\right)\)

            \(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x+5y-10\right)\)

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

4) \(ax-bx-a^2+2ab-b^2=x\left(a-b\right)-\left(a^2-2ab+b^2\right)\)

                            \(=x\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(x-a+b\right)\)

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

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

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

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

Phân tích các đa thức sau thành nhân tử :

1) x^3 + x^2y - 4x - 4y

2) x^3 - 3x^2 +1 - 3x

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

4) x^2 - 2x - 15

5) 2x^2 +3x - 5

6) 2x^2 - 18

7) x^2 - 7xy + 10y^2

8) x^3 - 2x^2 + x - xy^2

Làm nhanh giúp mình với nhé .....mình đang cần gấp[[[[

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\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)

\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)

\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)

Đặt ....

đúng ,mình k 2 nhé

a) Ta có: \(x^2\left(x-1\right)+16\left(1-x\right)\)

\(=x^2\left(x-1\right)-16\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-16\right)\)

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

b) Ta có: \(5x^2-5y^2-10x+10y\)

\(=5\left(x^2-y^2\right)-10\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\cdot2\)

\(=5\left(x-y\right)\left(x+y-2\right)\)

c) Ta có: \(x^2+4x-y^2+4\)

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

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

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

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

\(=-xy\left(3y+xy-5x\right)\)

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

\(=3y^3+6y+3+xy-x\)

Xem lại nhé ko phân tích được

\(12xy^2-12xy+3x\)

\(=3x\left(4y^2-4y+1\right)\)

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

\(10x^2\left(x+y\right)-5\left(2x+2y\right)y^2\)

\(=10x^2\left(x+y\right)-10\left(x+y\right)y^2\)

\(=10\left(x+y\right)\left(x-y\right)\left(x+y\right)\)

\(=10\left(x+y\right)^2\left(x-y\right)\)

Phân tích đa thức thành nhân tử:

a) \(3a^2-3ab+9b-9a=3a\left(a-b\right)+9\left(b-a\right)=3\left(a-b\right)\left(a-3\right)\)

b) \(2xm^3-2m=2m\left(xm^2-1\right)\)

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

Tìm x:

a) \(8x^2+10x+3=0\)

\(\Leftrightarrow8x^2+12x-2x-3=0\Leftrightarrow4x\left(2x+3\right)-\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{1}{4}\end{array}\right.\)

b) \(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)