Bài 6: Phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung

a)

\(=2y\left(10y-3\right)+15\left(10y-3\right)\\ =\left(10y-3\right)\left(2y+15\right)\)

b)

\(=3x\left(x-7y\right)-24\left(x-7y\right)\\ =\left(x-7y\right)\left(3x-24\right)\)

\(=3\left(x-7y\right)\left(x-8\right)\)

c)

\(=21x\left(4x+5\right)+9\left(4x+5\right)\\ =\left(4x+5\right)\left(21x+9\right)\\ =3\left(4x+5\right)\left(7x+3\right)\)

d)

\(=12\left(x-5\right)+5y\left(5-x\right)\\ =12\left(x-5\right)-5y\left(x-5\right)\\ =\left(x-5\right)\left(12-5y\right)\)

e)

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

f)

\(=2xy\left(y-3z\right)-10x\left(3z-y\right)\\ =2xy\left(y-3z\right)+10x\left(y-3z\right)\\ =\left(y-3z\right)\left(2xy+10x\right)\\ =2x\left(y-3z\right)\left(y+5\right)\)

`HaNa☘D`

a) \(2y\left(10y-3\right)-15\left(3-10y\right)\)

\(=2y\left(10y-3\right)+15\left(10y-3\right)\)

\(=\left(10y-3\right)\left(2y-15\right)\)

b) \(3x\left(x-7y\right)+24\left(7y-x\right)\)

\(=3x\left(x-7y\right)-24\left(x-7y\right)\)

\(=3\left(x-7y\right)\left(x-8\right)\) 

c) \(21x\left(4x+5\right)+36x+45\)

\(=21x\left(4x+5\right)+9\left(4x+5\right)\)

\(=3\left(7x+3\right)\left(4x+5\right)\)

d) \(12x-60+5y\left(5-x\right)\)

\(=12\left(x-5\right)-5y\left(x-5\right)\)

\(=\left(x-5\right)\left(12-5y\right)\)

e) \(5x\left(2y-z\right)+2x^2y+x^2z\)

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

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

f) \(2xy^2-6xyz-10x\left(3z-y\right)\)

\(=2xy\left(y-3z\right)+10x\left(y-3z\right)\)

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

(x+y)^2=(x-y)^2+4xy

=8^2+4*20

=64+80

=144

1) 4x³ - 16x

= 4x(x² - 4)

= 4x(x - 2)(x + 2)

2) x⁴ - y⁴

= (x² - y²)(x² + y²)

= (x - y)(x + y)(x² + y²)

3) x² - 6x + 9 - y²

= (x² - 6x + 9) - y²

= (x - 3)² - y²

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

4) x² - 4x - 5

= x² - 4x + 4 - 9

= (x² - 4x + 4) - 3²

= (x - 2)² - 3²

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

= (x - 5)(x + 1)

5) x² - 9x + 20

= x² - 4x - 5x + 20

= (x² - 4x) - (5x - 20)

= x(x - 4) - 5(x - 4)

= (x - 4)(x - 5)

6) xy² + x²y + 1/4 y³

= y(x² + xy + 1/4 y²)

= y(x + y/2)²

7) x³ - y⁶

= x³ - (y²)³

= (x - y²)(x² + xy² + y⁴)

8) x² + 2xy - 4x - 8y

= (x² + 2xy) - (4x + 8y)

= x(x + 2y) - 4(x + 2y)

= (x + 2y)(x - 4)

9) 3x² + 3x - 6

= 3(x² + x - 6)

= 3(x² - 2x + 3x - 6)

= 3[(x² - 2x) + (3x - 6)]

= 3[x(x - 2) + 3(x - 2)]

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

10) 3x² - 4x - 7

= 3x² + 3x - 7x - 7

= (3x² + 3x) - (7x - 7)

= 3x(x + 1) - 7(x + 1)

= (x + 1)(3x - 7)

11) Ta có:

49y² + 28y + 4

= (7y)² + 2.7y.2 + 2²

= (7y + 2)²

Vậy độ dài cạnh hình vuông là 7y + 2

\(-x^2-4x-5\\ =-\left(x^2+4x+5\right)\\ =-\left(x^2+4x+4+1\right)\\ =-\left(x^2+2.2x+2^2+1\right)\\ =-\left(x^2+2.2x+2^2\right)-1\\ =-\left(x+2\right)^2-1\le-1\)

Giá trị của biểu thức \(-x^2-4x-5\) đạt max khi và chỉ khi \(-\left(x+2\right)^2=0\Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

`HaNa☘D`

-x^2-4x-5

=-(x^2+4x+5)

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

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

Dấu = xảy ra khi x=-2

a) \(16x^2-1\)

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

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

b) \(\left(x+2\right)^2-49y^2\)

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

\(=\left[\left(x+2\right)-7y\right]\left[\left(x+2\right)+7y\right]\)

\(=\left(x+2-7y\right)\left(x+2+7y\right)\)

c) \(4x^2-12xy+9y^2\)

\(=\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2\)

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

d) \(\left(a+b\right)^2-\left(2a-b\right)^2\)

\(=\left[\left(a+b\right)+\left(2a-b\right)\right]\left[\left(a+b\right)-\left(2a-b\right)\right]\)

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

\(=3a\cdot\left(2b-a\right)\)

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

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

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

g) \(-3x^2+6xy-3y^2\)

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

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

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

a: 16x^2-1=(4x)^2-1=(4x-1)(4x+1)

b: (x+2)^2-49y^2

=(x+2)^2-(7y)^2

=(x+2+7y)(x+2-7y)

c: 4x^2-12xy+9y^2=(2x-3y)^2

d: (a+b)^2-(2a-b)^2

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

=(2b-a)*3a

g: =-3(x^2-2xy+y^2)

=-3(x-y)^2

\(a,6a^2b+9ab^2\)

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

\(b,5x^3y^2-15x^2y^3\)

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

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

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

\(d,\left(x-y\right)^2-x\left(x-y\right)\)

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

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

\(e,y\left(x-1\right)-x\left(1-x\right)\)

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

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

\(g,2a\left(a-b\right)+2b\left(b-a\right)\)

\(=2a\left(a-b\right)-2b\left(a-b\right)\)

\(=\left(2a-2b\right)\left(a-b\right)\)

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

#Urushi☕

a: 6a^2b+9ab^2

=3ab*2a+3ab*3b

=3ab(2a+3b)

b: 5x^3y^2-15x^2y^3

=5x^2y^2*x-5x^2y^2*3y

=5x^2y^2(x-3y)

c: 2x(x+1)-3y(x+1)

=(x+1)(2x-3y)

d: =(x-y)(x-y-x)

=-y(x-y)

e: =y(x-1)+x(x-1)

=(x-1)(x+y)

g: =2a(a-b)-2b(a-b)

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

=2(a-b)^2

a)\(6a^2b+9ab^2\)=ab(6a+9b)

b)\(5x^3y^2-15x^2y^3\)=5\(x^2y^2\)(x-3y)

c)2x(x+1)-3y(x+1)=(x+1)(2x-3y)

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

e)y(x-1)-x(1-x)=(x-1)(y+x)

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

`#040911`

`7x - 6x^2 - 2`

`= -6x^2 + 3x + 4x - 2`

`= -(6x^2 - 3x) +  (4x - 2)`

`= -3x(2x - 1) + 2(2x - 1)`

`= (-3x + 2)(2x - 1)`

=-6x^2+3x+4x-2

=-3x(2x-1)+2(2x-1)

=(2x-1)(-3x+2)

Bài 5:

a)

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

b)

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

c)

\(3x^2-12y^2\\ =\left(\sqrt{3}x\right)^2-\left(\sqrt{12}y\right)^2\\ =\left(\sqrt{3}x-\sqrt{12}y\right)\left(\sqrt{3}x+\sqrt{12}y\right)\\ =\sqrt{3}\left(x-\sqrt{4}y\right).\sqrt{3}\left(x+\sqrt{4}y\right)\\ =3\left(x-\sqrt{4}y\right)\left(x+\sqrt{4}y\right)\)

7:

a: (2x-1)^2-25=0

=>(2x-1)^2=25

=>2x-1=-5 hoặc 2x-1=5

=>2x=6 hoặc 2x=-4

=>x=-2 hoặc x=3

b: 8x^3-50x=0

=>4x^3-25x=0

=>x(4x^2-25)=0

=>x(2x-5)(2x+5)=0

=>x=0 hoặc 2x-5=0 hoặc 2x+5=0

=>x=0;x=5/2;x=-5/2

c: 3x(x-1)+(x-1)=0

=>(x-1)(3x+1)=0

=>x=1 hoặc x=-1/3

d: =>2(x+3)-x(x+3)=0

=>(x+3)(2-x)=0

=>x=-3 hoặc x=2

e: Thiếu vế phải rồi bạn

f: x^3+27+(x+3)(x-9)=0

=>(x+3)(x^2-3x+9)+(x+3)(x-9)=0

=>(x+3)(x^2-3x+9+x-9)=0

=>(x+3)(x^2-2x)=0

=>x(x-2)(x+3)=0

=>\(x\in\left\{0;2;-3\right\}\)

Bài 6: 

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

Thay x = 3 vào biểu thức B, ta có:

\(B=\left(3-1\right)\left(3-2\right)^2=2\cdot1^2=2\)

Đề:Phân tích bằng phương pháp tách một hạng tử thành nhiều hạng tử

mọi người giúp em với ạa:>

10: =x^2-3x-4x+12

=x(x-3)-4(x-3)

=(x-3)(x-4)

11: =x^2-2x-5x+10

=x(x-2)-5(x-2)

=(x-2)(x-5)

12: =x^2+x+5x+5

=x(x+1)+5(x+1)

=(x+1)(x+5)

13: =3x^2-6x+x-2

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

=(x-2)(3x+1)

14: =2x^2+4x-3x-6

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

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

15: =7x^2+49x+x+7

=7x(x+7)+(x+7)

=(x+7)(7x+1)

18: =2x^2+4x+x+2

=2x(x+2)+(x+2)

=(x+2)(2x+1)

20: =2(x^2+5x+4)

=2(x+1)(x+4)

Bài 1A:

\(a,x^2-xy+x-y\\ =x\left(x-y\right)+\left(x-y\right)\\ =\left(x+1\right)\left(x-y\right)\\ b,xz+yz+4x+4y\\ =z\left(x+y\right)+4\left(x+y\right)\\ =\left(x+y\right)\left(z+4\right)\\ c,x^2-x-y^2+y\\ =\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x+y-1\right)\\ d,x^2+2x+2z-z^2\\ =\left(x-z\right)\left(x+z\right)+2\left(x+z\right)\\ =\left(x+z\right)\left(x-z+2\right)\)

Bài 1B:

\(a,x^2+2xy+x+2y\\ =x\left(x+2y\right)+\left(x+2y\right)\\ =\left(x+2y\right)\left(x+1\right)\\ b,2xy+yz+2x+z\\ =y\left(2x+z\right)+\left(2x+z\right)\\ =\left(y+1\right)\left(2x+z\right)\\ c,y^2-2y-z^2-2z\\ =\left(y-z\right)\left(y+z\right)-2\left(y+z\right)\\ =\left(y+z\right)\left(y-z-2\right)\\ d,x^3-x-y+y^3\\ =\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2-1\right)\)

1B

a)

\(=x\left(x+2y\right)+\left(x+2y\right)\\ =\left(x+1\right)\left(x+2y\right)\)

b)

\(=y\left(2x+z\right)+\left(2x+z\right)\\ =\left(y+1\right)\left(2x+z\right)\)

c)

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

d)

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