\(a,4x^2-12xy+9y^2=\left(2x-3y\right)^2\)

\(b,x^2-9x+20=x^2-4x-5x+20\)

\(=x\left(x-4\right)-5\left(x-4\right)\)

\(=\left(x-4\right)\left(x-5\right)\)

\(c,x^2+7x+12=x^2+3x+4x+12\)

\(=x\left(x+3\right)+4\left(x+3\right)\)

\(=\left(x+3\right)\left(x+4\right)\)

\(a,3x\left(x+1\right)+9\left(x+1\right)\\ =\left(3x+9\right)\left(x+1\right)\\ =3\left(x+3\right)\left(x+1\right)\\ b,x^2-5xy+2x-10y\\ =x\left(x-5y\right)+2\left(x-5y\right)\\ =\left(x+2\right)\left(x-5y\right)\)

BÀI 2: Phân tích thành nhân tử .

`a, 3x(x+1)+9(x+1)`

`= (x+1)(3x+9)`

`=3(x+1)(x+3)`

`b,x^2-5xy+2x-10y`

`=x(x-5y)+2(x-5y)`

`=(x-5y)(x+2)`

\(a,=x-x^2=x\left(1-x\right)\\ b,=x^2+3x-2x-6=\left(x+3\right)\left(x-2\right)\\ c,=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)

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

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

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

(x - 4)(x² + 4x + 16) - x(x² - 6) = 2

x³ - 64 - x³ + 6x = 2

6x = 2 + 64

6x = 66

x = 66 : 6

x = 11

x³ - 27 + 3x(x - 3)

= (x - 3)(x² + 3x + 9) + 3x(x - 3)

= (x - 3)(x² + 3x + 9 + 3x)

= (x - 3)(x² + 6x + 9)

= (x - 3)(x + 3)²

5x³ - 7x² + 10x - 14

= 5x(x² + 2) - 7(x² + 2)

= (x² + 2)(5x - 7)

C

b: \(\left(x^2+4\right)^2-16x^2\)

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

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

c: \(x^5-x^4+x^3-x^2\)

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

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

Lời giải:

a. Bạn xem lại đề

b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)

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

c.

\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)

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

a) 7x.(-y)+2(y-x)2

=>-7xy+4y-4x

b)(x^2+4)-16x^2

=>x^2+4-16x^2

=>-15x^2+4

c)x^5-x^4+x^3-x^2

=>x^4(x-1)+x^2(x-1)

=>(x^4+x^2)(x-1)

b, A=[(a+1)(a+7)][(a+3)(a+5)]+15

=>A=(a²+8a+7)(a²+8a+15)+15

Đặt a²+8a+11= t

=>a²+8a+7= t-4 và a²+8a+15= t+4

=>A=(t-4)(t+4)+15

=>A=t²-16+15

      =t²-1=(t-1)(t+1)

Thay t = a²+8a+11

=>A=(a²+8a+11-1)(a²+8a+11+1)

=>A=(a²+8a+10)(a²+8a+12)

a)   \(x^2+2xy+7x+7y+y^2+10\)

\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)

\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)

\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)

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

c,=(X⁵ - X²)+(X² +X+1)

  =X²(X³-1)+(X²+X+1)

  =X²(X-1)(X²+X+1)+(X²+X+1)

=(X³-X²+1)(X²+X+1)

\(5-7x^2=\left(\sqrt{5}\right)^2-\left(x\sqrt{7}\right)^2\)

\(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)

\(3+4x=\left(\sqrt{3}\right)^2-\left(2\sqrt{x}\right)^2\) ( do x<0 )

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

Bài 2: 

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

b: \(5x^2+5xy-x-y\)

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

\(=\left(x+y\right)\left(5x-1\right)\)

c:\(-6x^2+7x-2\)

\(=-6x^2+3x+4x-2\)

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

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

1.

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

b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

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

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

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

2.

a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

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

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)