Question

Giải giúp em hết với, mai em cần nộp gấp, cảm ơn ạ!

 

Answer 1

1a.

$x^2-5x+6=x^2-2x-(3x-6)=x(x-2)-3(x-2)=(x-2)(x-3)$

1b.

$3x^2+9x-30=3(x^2+3x-10)=3(x^2-2x+5x-10)$

$=3[x(x-2)+5(x-2)]=3(x-2)(x+5)$

1c.
$x^2-3x+2=(x^2-x)-(2x-2)=x(x-1)-2(x-1)=(x-1)(x-2)$

1d.

$x^2-9x+18=x^2-3x-(6x-18)=x(x-3)-6(x-3)=(x-3)(x-6)$

1e.

$x^2-6x+8=x^2-2x-(4x-8)=x(x-2)-4(x-2)=(x-2)(x-4)$

Answer 2

1f.
$x^2-5x-14=x^2-7x+2x-14=x(x-7)+2(x-7)=(x+2)(x-7)$

1g.

$x^2+6x+5=(x^2+x)+(5x+5)=x(x+1)+5(x+1)=(x+1)(x+5)$

1h.

$x^2-7x+12=x^2-3x-(4x-12)=x(x-3)-4(x-3)=(x-3)(x-4)$

1i.

$x^2-7x+10=(x^2-2x)-(5x-10)=x(x-2)-5(x-2)=(x-2)(x-5)$

Answer 3

2a.

$3x^2-5x-2=(3x^2-6x)+(x-2)=3x(x-2)+(x-2)$

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

2b. 

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

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

2c.

$7x^2+50x+7=(7x^2+49x)+(x+7)$

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

2d. 

$12x^2+7x-12=(12x^2-9x)+(16x-12)$

$=3x(4x-3)+4(4x-3)=(4x-3)(3x+4)$

2e.

$15x^2+7x-2=15x^2-3x+10x-2$

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

Answer 4

2f.

$a^2-5a-14=a^2-7a+2a-14=a(a-7)+2(a-7)$

$=(a-7)(a+2)$

2g.

$2m^2+10m+8=2(m^2+5m+4)$

$=2(m^2+m+4m+4)=2[m(m+1)+4(m+1)]$

$=2(m+1)(m+4)$

2h.

$4p^2-36p+56=4(p^2-9p+14)=4[p^2-2p-(7p-14)]$

$=4[p(p-2)-7(p-2)]=4(p-2)(p-7)$

2i.

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

$=x(2x+1)+2(2x+1)=(2x+1)(x+2)$
 

Answer 5

Bài 3:

a. $x^2+4xy-21y^2=x^2+4xy+4y^2-25y^2$
$=(x+2y)^2-(5y)^2=(x+2y-5y)(x+2y+5y)$

$=(x-3y)(x+7y)$

b.

$5x^2+6xy+y^2=9x^2+6xy+y^2-4x^2$

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

c.

$x^2+2xy-15y^2=x^2+2xy+y^2-16y^2$

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

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

d.

$(x-y)^2+4(x-y)-12=(x-y)^2+4(x-y)+4-16$

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

$=(x-y-2)(x-y+6)$

e. 

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

$=x^2-2xy-(5xy-10y^2)$

$=x(x-2y)-5y(x-2y)=(x-2y)(x-5y)$

f. 

$x^2yz+5xyz-14yz=yz(x^2+5x-14)$

$=yz(x^2-2x+7x-14)$

$=yz[x(x-2)+7(x-2)]=yz(x-2)(x+7)$

Answer 6

4a.

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

$=(a^2+1)^2-a^2=(a^2+1-a)(a^2+1+a)$
4b. 

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

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

4c. 

$x^4+4x^2-5=(x^4+4x^2+4)-9=(x^2+2)^2-3^2$

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

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

Answer 7

4d. 

$x^3-19x-30=x^3+2x^2-(2x^2+4x)-(15x+30)$

$=x^2(x+2)-2x(x+2)-15(x+2)=(x+2)(x^2-2x-15)$

$=(x+2)[x^2+3x-(5x+15)]$

$=(x+2)[x(x+3)-5(x+3)]=(x+2)(x+3)(x-5)$

4e.

$x^3-7x-6=(x^3+1)-(7x+7)=(x+1)(x^2-x+1)-7(x+1)$

$=(x+1)(x^2-x+1-7)=(x+1)(x^2-x-6)$

$=(x+1)[x^2+2x-(3x+6)]$
$=(x+1)[x(x+2)-3(x+2)]=(x+1)(x+2)(x-3)$

4f.

$x^3-5x^2-14x=x(x^2-5x-14)=x[x^2+2x-(7x+14)]$

$=x[x(x+2)-7(x+2)]=x(x+2)(x-7)$

Answer 8

5a. $x^4+4=(x^2)^2+2^2+2.x^2.2-4x^2$

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

5b. $x^4+64=(x^2)^2+8^2+2.x^2.8-16x^2$

$=(x^2+8)^2-(4x)^2=(x^2+8-4x)(x^2+8+4x)$

5c. $x^8+x^7+1=(x^8-x^2)+(x^7-x)+(x^2+x+1)$

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

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

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

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

$=(x^2+x+1)[(x-1)(x^3+1)(x^2+x)+1]$

$=(x^2+x+1)(x^6-x^4+x^3-x+1)$

5d. $x^8+x^4+1=(x^8+x^7+1)-(x^7-x^4)$

$=(x^2+x+1)(x^6-x^4+x^3-x+1)-x^4(x^3-1)$

$=(x^2+x+1)(x^6-x^4+x^3-x+1)-x^4(x-1)(x^2+x+1)$

$=(x^2+x+1)(x^6-x^4+x^3-x+1-x^5+x^4)$

$=(x^2+x+1)(x^6-x^5+x^3-x+1)$

$=(x^2+x+1)[x^4(x^2-x+1)-(x^4-x^3+x^2)+(x^2-x+1)]$

$=(x^2+x+1)[x^4(x^2-x+1)-x^2(x^2-x+1)+(x^2-x+1)]$

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

5e. $x^5+x+1=(x^5-x^2)+(x^2+x+1)$

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

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

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

 

Answer 9

5f.

$x^3+x^2+4=(x^2+2x^2)-(x^2-4)$

$=x^2(x+2)-(x-2)(x+2)=(x+2)[x^2-(x-2)]=(x+2)(x^2-x+2)$

5g.

$x^4+2x^2-24=(x^4+2x^2+1)-25$

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

$=(x^2-4)(x^2+6)=(x-2)(x+2)(x^2+6)$

5h.

$x^3-2x-4=(x^3-2x^2)+(2x^2-4x)+(2x-4)$

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

5i.

$a^4+4b^4=(a^2)^2+(2b^2)^2+2.a^2.2b^2-4a^2b^2$

$=(a^2+2b^2)^2-(2ab)^2=(a^2+2b^2-2ab)(a^2+2b^2+2ab)$

Answer 10

6a.

Đặt $x^2+x=a$ thì:

$(x^2+x)^2-14(x^2+x)+24$

$=a^2-14a+24=a^2-2a-(12a-24)$

$=a(a-2)-12(a-2)=(a-12)(a-2)=(x^2+x-12)(x^2+x-2)$

$=(x^2-3x+4x-12)(x^2-x+2x-2)$

$=[x(x-3)+4(x-3)][x(x-1)+2(x-1)]=(x-3)(x+4)(x-1)(x+2)$

6b. Đặt $x^2+x=a$ thì:

$(x^2+x)^2+4x^2+4x-12$

$=(x^2+x)^2+4(x^2+x)-12$

$=a^2+4a-12=(a^2-2a)+(6a-12)$

$=a(a-2)+6(a-2)=(a+6)(a-2)=(x^2+x+6)(x^2+x-2)$

$=(x^2+x+6)(x^2-x+2x-2)$

$=(x^2+x+6)[x(x-1)+2(x-1)]=(x^2+x+6)(x-1)(x+2)$

6c.

$x^4+2x^3+5x^2+4x-12$

$=(x^4+2x^3+x^2)+4(x^2+x)-12$
$=(x^2+x)^2+4(x^2+x)-12$

$=(x^2+x+6)(x-1)(x+2)$ (giống phần b)

Answer 11

6d.

$(x+1)(x+2)(x+3)(x+4)+1$

$=(x+1)(x+4)(x+2)(x+3)+1$

$=(x^2+5x+4)(x^2+5x+6)+1$

$=a(a+2)+1$ (đặt $x^2+5x+4=a$)

$=a^2+2a+1$

$=(a+1)^2=(x^2+5x+5)^2$

6e.

$(x+1)(x+3)(x+5)(x+7)+15$

$=(x+1)(x+7)(x+3)(x+5)+15$

$=(x^2+8x+7)(x^2+8x+15)+15$

$=a(a+8)+15$ (đặt $x^2+8x+7=a$)

$=a^2+8a+15=a^2+3a+5a+15$

$=a(a+3)+5(a+3)=(a+3)(a+5)=(x^2+8x+10)(x^2+8x+12)$

$=(x^2+8x+10)(x^2+2x+6x+12)$

$=(x^2+8x+10)[x(x+2)+6(x+2)]$

$=(x^2+8x+10)(x+2)(x+6)$

6f.

$(x+1)(x+2)(x+3)(x+4)-24$

$=(x+1)(x+4)(x+2)(x+3)-24$

$=(x^2+5x+4)(x^2+5x+6)-24$

$=a(a+2)-24$ (đặt $x^2+5x+4=a$)

$=a^2+2a-24=a^2-4a+6a-24$

$=a(a-4)+6(a-4)=(a-4)(a+6)$

$=(x^2+5x)(x^2+5x+10)=x(x+5)(x^2+5x+10)$

Answer 12

7a.

Đặt $x^2+4x+8=a$ thì:

$(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$

$=a^2+3ax+2x^2=(a^2+ax)+(2ax+2x^2)$

$=a(a+x)+2x(a+x)=(a+2x)(a+x)$

$=(x^2+6x+8)(x^2+5x+8)$

$=(x^2+2x+4x+8)(x^2+5x+8)$

$=[x(x+2)+4(x+2)](x^2+5x+8)=(x+4)(x+2)(x^2+5x+8)$

7b. Đặt $x^2+x+1=a$ thì:

$(x^2+x+1)(x^2+x+2)-12$

$=a(a+1)-12=a^2+a-12$

$=(a^2-3a)+(4a-12)=a(a-3)+4(a-3)$

$=(a+4)(a-3)=(x^2+x+5)(x^2+x-2)$

$=(x^2+x+5)[(x^2-x)+(2x-2)]$

$=(x^2+x+5)[x(x-1)+2(x-1)]=(x^2+x+5)(x+2)(x-1)$

Answer 13

7c.

Đặt $x^2+8x+7=a$ thì:

$(x^2+8x+7)(x^2+8x+15)+15$

$=a(a+8)+15$

$=a^2+8a+15=a^2+3a+5a+15$

$=a(a+3)+5(a+3)=(a+5)(a+3)$

$=(x^2+8x+12)(x^2+8x+10)$

$=(x^2+2x+6x+12)(x^2+8x+10)$
$=[x(x+2)+6(x+2)](x^2+8x+10)=(x+2)(x+6)(x^2+8x+10)$

7d.

$(x+2)(x+3)(x+4)(x+5)-24$

$=(x+2)(x+5)(x+3)(x+4)-24$

$=(x^2+7x+10)(x^2+7x+12)-24$

$=a(a+2)-24$ (đặt $x^2+7x+10=a$)

$=a^2+2a-24$

$=a^2-4a+6a-24=a(a-4)+6(a-4)$
$=(a-4)(a+6)=(x^2+7x+6)(x^2+7x+16)$

$=[(x^2+x)+(6x+6)](x^2+7x+16)$

$=[x(x+1)+6(x+1)](x^2+7x+16)$

$=(x+1)(x+6)(x^2+7x+16)$

Answer 14

8a.

$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+2xyz$

$=(x^2y+xy^2+xyz)+(x^2z+xz^2+xyz)+(y^2z+yz^2)$

$=xy(x+y+z)+xz(x+z+y)+yz(y+z)$

$=(x+y+z)(xy+xz)+yz(y+z)=x(x+y+z)(y+z)+yz(y+z)$

$=(y+z)[x(x+y+z)+yz]=(y+z)[x(x+y)+z(x+y)]$

$=(y+z)(x+y)(x+z)$

8b.

$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+3xyz$

$=(x^2y+xy^2+xyz)+(x^2z+xz^2+xyz)+(y^2z+yz^2+xyz)$

$=xy(x+y+z)+xz(x+y+z)+yz(x+y+z)$

$=(x+y+z)(xy+xz+yz)$

8c.

$x(y^2-z^2)+y(z^2-x^2)+z(x^2-y^2)$

$=x(y^2-z^2)-y[(y^2-z^2)+(x^2-y^2)]+z(x^2-y^2)$

$=(y^2-z^2)(x-y)-(y-z)(x^2-y^2)$

$=(y-z)(y+z)(x-y)-(y-z)(x-y)(x+y)$

$=(y-z)(x-y)(y+z-x-y)=(y-z)(x-y)(z-x)$

Answer 15

8d.

$xy(x-y)-xz(x+z)+yz(2x-y+z)$

$=xy(x-y)-xz(x+z)+yz[(x-y)+(x+z)]$

$=xy(x-y)-xz(x+z)+yz(x-y)+yz(x+z)$

$=(x-y)(xy+yz)+(x+z)(yz-xz)$

$=y(x-y)(x+z)-z(x+z)(x-y)$

$=(x-y)(x+z)(y-z)$

8e.

$x(y+z)^2+y(z+x)^2+z(x+y)^2-4xyz$

$=xy^2+xz^2+yz^2+yx^2+zx^2+zy^2+2xyz$

$=xy(x+y+z)+xz(x+y+z)+yz(y+z)$

$=(x+y+z)(xy+xz)+yz(y+z)$

$=x(x+y+z)(y+z)+yz(y+z)=(y+z)(x^2+xy+xz+yz)$

$=(y+z)[x(x+y)+z(x+y)]=(y+z)(x+y)(x+z)$

8f.

$yz(y+z)+xz(z-x)-xy(x+y)$

$=yz(y+z)+xz(z-x)-xy[(y+z)-(z-x)]$

$=yz(y+z)+xz(z-x)-xy(y+z)+xy(z-x)$

$=(y+z)(yz-xy)+(z-x)(xz+xy)$

$=(y+z)y(z-x)+(z-x)x(z+y)$

$=(y+z)(z-x)(y+x)$

Answer 16

Bài 9: 

Ta thấy:

$B=x^3+6x^2-19x-24=(x^3-x)+6x^2-18x-24$

$=x(x^2-1)+6(x^2-3x-4)$

$=x(x-1)(x+1)+6(x^2-3x-4)$
Vì $x(x-1)$ là tích 2 số nguyên liên tiếp nên $x(x-1)\vdots 2$

$\Rightarrow x(x-1)(x+1)\vdots 2$

Mặt khác: $x(x-1)(x+1)$ là tích 3 số nguyên liên tiếp nên $x(x-1)(x+1)\vdots 3$

Mà $(2,3)=1$ nên $x(x-1)(x+1)\vdots 6$

$\Rightarrow B=x(x-1)(x+1)+6(x^2-3x-4)\vdots 6$

Ta có đpcm.

Answer 17

Bài 10:

$y^2+2(x^2+1)=2y(x+1)$

$\Leftrightarrow y^2+2x^2+2-2xy-2y=0$

$\Leftrightarrow (y^2+x^2-2xy)+2(x-y)+1+x^2-2x+1=0$

$\Leftrightarrow (x-y)^2+2(x-y)+1+(x^2-2x+1)=0$

$\Leftrightarrow (x-y+1)^2+(x-1)^2=0$

$\Rightarrow (x-y+1)^2=(x-1)^2=0$

$\Leftrightarrow x=1; y=2$

Answer 18

Bài 11:
$3x^2-2y^2-5xy+x-2y-7=0$

$\Leftrightarrow 3x^2+x(1-5y)-(2y^2+2y+7)=0$

Coi đây là pt bậc 2 ẩn $x$

$\Delta=(1-5y)^2+12(2y^2+2y+7)=49y^2+14y+85$

$=(7y+1)^2+84$

Để pt có nghiệm nguyên thì:

$(7y+1)^2+84=t^2$ với $t$ là số tự nhiên 

$\Leftrightarrow 84=(t-7y-1)(t+7y+1)$

Vì $t-7y-1, t+7y+1$ cùng tính chẵn lẻ nên ta xét các TH sau:

$t+7y+1=2; t-7y-1=42$

$t+7y+1=-2; t-7y-1=-42$

$t+7y+1=42; t-7y-1=2$

$t+7y+1=-42; t-7y-1=-2$

$t+7y+1=6; t-7y-1=14$

$t+7y+1=-6; t-7y-1=-14$

$t+7y+1=-14; t-7y-1=-6$

$t+7y+1=14; t-7y-1=6$

Đến đây thì đơn giản rồi.