Phân tích đa thức thành nhân tử:

\(6a^2b-18ab\)

\(=6ab\left(a-3\right)\)

6a²b-18ab = 6ab ( a-3)

\(5a^2-14ab-3b^2\\ =5a^2-15ab-ab-3b^2\\ =5a\left(a-3b\right)-b\left(a-3b\right)\\ =\left(5a-b\right)\left(a-3b\right)\)

\(5a^2-14ab-3b^2\)

\(=5a^2-15ab+ab-3b^2\)

\(=5a\left(a-3b\right)+b\left(a-3b\right)\)

\(=\left(a-3b\right)\left(5a+b\right)\)

a: \(3ab-6a^2b\)

\(=3ab\cdot1-3ab\cdot2a\)

=3ab(1-2a)

b: \(x^3-6x\)

\(=x\cdot x^2-x\cdot6\)

\(=x\left(x^2-6\right)\)

c: \(x^2-y^2-9x+9y\)

\(=\left(x^2-y^2\right)-\left(9x-9y\right)\)

\(=\left(x-y\right)\left(x+y\right)-9\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-9\right)\)

d: \(5x^2+10xy+5y^2\)

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

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

giải bài toán: cho tam giác MNP, NTlà phân giác của góc N biết MN=4cm, NT=10cm, MP=8cm:TínhTM, TP? 

a) 10x³-90x⁵=10x³(1-9x²)=10x³(1-3x)(1+3x)

b)3a²-6a²b+5a-10ab=(3a²-6a²b)+(5a-10ab)=3a²(1-2b)+5a(1-b)=(3a²+5a)(1-2b)=a(3a+5)(1-2b)

c) 7a³-28a²+28a=7a(a²-4a+4)

d) 45a³-30a²+5a-500=5(9a³-6a²+a-100)

1: 2a+2b=2(a+b)

2: 2a+4b+6c

=2*a+2*2b+2*3c

=2(a+2b+3c)

3: \(-7a-14ab-21b=-7\left(a+2ab+3b\right)\)

4: \(2ax-2ay+2a=2a\left(x-y+1\right)\)

5: \(=3a\cdot ax-3a\cdot2ay+3a\cdot4=3a\left(ax-2ay+4\right)\)

6: \(=2\cdot2ax-2\cdot ay-2\cdot1=2\cdot\left(2ax-ay-1\right)\)

7: =a^2-(2b)^2

=(a-2b)(a+2b)

8: =(5a)^2-1^2

=(5a-1)(5a+1)

9: =9(16a^2-9)

=9(4a-3)(4a+3)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)