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

Bài 16
a)  \(=x^2-2x-3x+6\)

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

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

c) \(=x^2-x-2x+2\)

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

d) \(=x^2-3x-6x+18\)

\(=x\left(x-3\right)-6\left(x-3\right)=\left(x-6\right)\left(x-3\right)\)

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

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

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

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

f) \(=x^2+2x-7x-14=x\left(x+2\right)-7\left(x+2\right)=\left(x-7\right)\left(x+2\right)\)

a: =(x^3-1)+5(x^2-1)+3(x-1)

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

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

=(x-1)(x^2+6x+9)

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

b: =a^4(a+1)+a^2(a+1)+(a+1)

=(a+1)(a^4+a^2+1)

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

=(a+1)[(a^2+1)^2-a^2]

=(a+1)(a^2+1-a)(a^2+1+a)

c: =(x-1)^3-y^3

=(x-1-y)[(x-1)^2+y(x-1)+y^2]

=(x-y-1)(x^2-2x+1+xy-y+y^2)

d: =x^2(5x-3y)-9y^2(5x-3y)

=(5x-3y)(x^2-9y^2)

=(5x-3y)(x-3y)(x+3y)

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

\(c,3x^4y^2+3x^3y^2+3xy^2+3y^2\\=3y^2\left(x^4+x^3+x+1\right)\\ =3y^2\left(x+1\right)^2\left(x^2-x+1\right) \\ d,4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\\ =-4\left(a^2-2ay+y^2\right)+4\left(x^2-2x+1\right)\\ =4\left(x-1\right)^2-4\left(a-y\right)^2\\ =4\left(x+y-a-1\right)\left(x-y+a-1\right)\)

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

a: =x^2+2xy+y^2-4x^2y^2

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

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

b: =(x^3-y^3)(x^3+y^3)

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

c: =25-(a^2-2ab+b^2)

=25-(a-b)^2

=(5-a+b)(5+a-b)

d: =(2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2)

=[a^2-(b^2-2bc+c^2)][(b+c)^2-a^2]

=(a-b+c)(a+b-c)(b+c-a)(b+c+a)

\(a,8x^3-64\\ =8\left(x^3-8\right)\\ =8\left(x-2\right)\left(x^2+2x+4\right)\\ b,1+8x^6y^3\\ =\left(2x^2y+1\right)\left(4x^4y^2-2x^2y+1\right)\\ c,125x^3+1\\ =\left(5x+1\right)\left(25x^2+5x+1\right)\\ d,8x^3-27\\ =\left(2x-3\right)\left(4x^2+6x+9\right)\\ e,27x^3+\dfrac{y^3}{8}\\ =\left(3x+\dfrac{y}{2}\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{y^2}{4}\right)\\ f,125x^3+27y^3\\ =\left(5x+3y\right)\left(25x^2+15xy+9y^2\right)\)

B=x^2-12x+6^2-8

=(x-6)^2-8

Biểu thức này ko thể luôn dương nha bạn

6:

A=n^4-2n^3-n^2+2n

=n^3(n-2)-n(n-2)

=(n-2)(n^3-n)

=n(n-1)(n+1)(n-2)

Vì n-2;n-1;n;n+1 là 4 số liên tiếp

nên n(n-1)(n+1)(n-2) chia hết cho 4!=24

5:

\(P=\dfrac{1}{2}\cdot48\left(7^2+1\right)\left(7^4+1\right)\cdot...\cdot\left(7^{16}+1\right)\)

\(=\dfrac{1}{2}\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\cdot...\cdot\left(7^{16}+1\right)\)

\(=\dfrac{1}{2}\left(7^4-1\right)\left(7^4+1\right)\cdot...\cdot\left(7^{16}+1\right)\)

\(=\dfrac{1}{2}\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

\(=\dfrac{1}{2}\left(7^{32}-1\right)\)

3:

a: 5m^3-45m

=5m(m^2-9)

=5m(m-3)(m+3)

b: x^4-16y^4

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

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

c: -x^2y+xy^2+1/4x^3

=x(1/4x^2-xy+y^2)

=x(1/2x-y)^2

d: x^2+6x-t^2+9

=(x^2+6x+9)-t^2

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

=(x+3+t)(x+3-t)

e: a^2-ab+a-b

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

=(a-b)(a+1)

g: x^3-2x^2-x+2

=x^2(x-2)-(x-2)

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

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

`a^3b^2 -4a+a^2b^3-4b`

`=(a^3b^2 +a^2b^3)-(4a+4b)`

`= a^2b^2(a+b) -4(a+b)`

`=(a+b)(a^2b^2-4)`