Bài 3:

a: \(10x\left(x-y\right)-8y\cdot\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=2\left(5x+4y\right)\left(x-y\right)\)

b: \(5x\left(x-2000\right)-x+2000\)

\(=5x\left(x-2000\right)-\left(x-2000\right)\)

\(=\left(x-2000\right)\left(5x-1\right)\)

Bài 4:

a: \(\left(a-b\right)\cdot x+\left(b-a\right)y-b+a\)

\(=\left(a-b\right)\cdot x-\left(a-b\right)\cdot y+a-b\)

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

b: \(\left(a+b-c\right)\cdot x^2-\left(c-a-b\right)\cdot x\)

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

\(=x\left(a+b-c\right)\left(x+1\right)\)

a: \(5x^2\left(x-2y\right)-15xy\left(x-2y\right)\)

\(=\left(x-2y\right)\left(5x^2-15xy\right)\)

\(=\left(5x\cdot x-5x\cdot3y\right)\left(x-2y\right)\)

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

b: \(x\left(x+y\right)+4x+4y\)

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

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

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

a: \(x^2-3x\)

\(=x\cdot x-x\cdot3\)

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

b: \(12x^3-6x^2+3x\)

\(=3x\cdot4x^2-3x\cdot2x+3x\cdot1\)

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

c: \(\dfrac{2}{5}x^2+5x^3+x^2y\)

\(=x^2\cdot\dfrac{2}{5}+x^2\cdot5x+x^2\cdot y\)

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

d: \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)

\(=7xy\left(2x-3y+4xy\right)\)

VD3:

b: \(4a^2-4b^2-4a+1\)

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

\(=\left(2a-1\right)^2-\left(2b\right)^2\)

\(=\left(2a-1-2b\right)\left(2a-1+2b\right)\)

c: \(4a^2-4b^2-4a+1\)

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

\(=\left(2a-1\right)^2-\left(2b\right)^2\)

\(=\left(2a-1-2b\right)\left(2a-1+2b\right)\)

d: \(a^3+6a^2+12a+8\)

\(=a^3+3\cdot a^2\cdot2+3\cdot a\cdot2^2+2^3\)

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

e: \(a^4+a^3+a^3b+a^2b\)

\(=\left(a^4+a^3\right)+\left(a^3b+a^2b\right)\)

\(=a^3\left(a+1\right)+a^2b\left(a+1\right)\)

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

f: \(a^3+3a^2+4a+12\)

\(=\left(a^3+3a^2\right)+\left(4a+12\right)\)

\(=a^2\left(a+3\right)+4\left(a+3\right)\)

\(=\left(a+3\right)\left(a^2+4\right)\)

g: \(a^3+4a^2+4a+3\)

\(=a^3+3a^2+a^2+3a+a+3\)

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

\(=\left(a+3\right)\left(a^2+a+1\right)\)

h: \(x^2-2x-3\)

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

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

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

h: \(x^2-10x+16\)

\(=x^2-2x-8x+16\)

\(=x\left(x-2\right)-8\left(x-2\right)\)

\(=\left(x-2\right)\left(x-8\right)\)

i: \(x^5-5x^3+4x\)

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

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

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

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

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

Lời giải:
ĐKXĐ: $x\neq 2; x\neq -3$

\(\frac{x+2}{x+3}-\frac{5}{(x+3)(x-2)}-\frac{1}{x-2}=\frac{(x+2)(x-2)-5-(x+3)}{(x+3)(x-2)}\\ =\frac{x^2-4-5-x-3}{(x+3)(x-2)}=\frac{x^2-x-12}{(x+3)(x-2)}\\ =\frac{(x+3)(x-4)}{(x+3)(x-2)}=\frac{x-4}{x-2}\)

bạn chụp lại đề được không bạn? Nó mờ quá á bạn

đây ah

Bài 4:

a. Ta thấy: $x^2-x+2=(x-\frac{1}{2})^2+1,75>0$ với mọi $x$.

Do đó để $B=\frac{x^2-x+2}{x-3}<0$ thì $x-3<0$

$\Leftrightarrow x<3$ 

b. 

$B=\frac{x(x-3)+2(x-3)+8}{x-3}=x+2+\frac{8}{x-3}$

Với $x$ nguyên, để $B$ nguyên thì $x-3$ phải là ước của 8.

$\Rightarrow x-3\in\left\{\pm 1; \pm 2; \pm 4; \pm 8\right\}$

$\Rightarrow x\in \left\{4; 2; 5; 1; -1; 7; 11; -5\right\}$

Bài 5:

\(\frac{\frac{x}{x-y}-\frac{y}{x+y}}{\frac{y}{x-y}+\frac{x}{x+y}}=\frac{\frac{x(x+y)-y(x-y)}{(x-y)(x+y)}}{\frac{y(x+y)+x(x-y)}{(x-y)(x+y)}}\)

\(=\frac{x(x+y)-y(x-y)}{y(x+y)+x(x-y)}=\frac{x^2+y^2}{x^2+y^2}=1\)

1) $x^2-y^2-2x+2y$

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

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

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

2) $2x+2y-x^2-xy$

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

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

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

3) $3a^2-6ab+3b^2-12c^2$

$=3(a^2-2ab+b^2-4c^2)$

$=3[(a^2-2ab+b^2)-4c^2]$

$=3[(a-b)^2-(2c)^2]$

$=3(a-b-2c)(a-b+2c)$

4) $x^2-25+y^2+2xy$

$=(x^2+2xy+y^2)-25$

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

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

5) $a^2+2ab+b^2-ac-bc$

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

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

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

6) $x^2-2x-4y^4-4y$

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

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

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

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

7) $x^2y-x^3-9y+9x$

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

$=x^2(y-x)-9(y-x)$

$=(y-x)(x^2-9)$

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

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

8) $x^2(x-1)+16(1-x)$

$=x^2(x-1)-16(x-1)$

$=(x-1)(x^2-16)$

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

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

9) $3x^2-6x+9x^3$

$=3x^2+3x-9x+9x^3$

$=3x(x+1)-9x(1-x^2)$

$=3x(x+1)+9x(x^2-1)$

$=3x(x+1)+9x(x-1)(x+1)$

$=(x+1)[3x+9x(x-1)]$

$=(x+1)(3x+9x^2-9x)$

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

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

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

10) $10x(x-y)-6y(y-x)$

$=10x(x-y)+6y(x-y)$

$=(x-y)(10x+6y)$

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

11) $3x^2+5y-3xy-5x$

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

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

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

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

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

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

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

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

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

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

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

14) $x^2-4x-5$

$=x^2+x-5x-5$

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

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

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

$=x^2+3x+5x+15$

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

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

16) $81x^4+4$

$=[(9x^2)^2+2\cdot9x^2\cdot 2+2^2]-2\cdot9x^2\cdot2$

$=(9x^2+2)^2-36x^2$

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

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

17) $2x^2+3x-5$

$=2x^2-2x+5x-5$

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

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

18) $16x-5x^2-3$

$=-5x^2+16x-3$

$=-5x^2+15x+x-3$

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

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

$Toru$

a: Xét ΔBAC có

D,M lần lượt là trung điểm của BA,BC

=>DM là đường trung bình của ΔBCA

=>DM//AC và \(DM=\dfrac{AC}{2}\)

DM//AC

E\(\in\)AC

Do đó: DM//AE

DM=AC/2

\(AE=\dfrac{AC}{2}\)

Do đó: DM=AE

Xét tứ giác ADME có

DM//AE

DM=AE

Do đó: ADME là hình bình hành

b: Để hình bình hành ADME trở thành hình chữ nhật thì \(\widehat{DAE}=90^0\)

=>\(\widehat{BAC}=90^0\)

c: Xét ΔABC có

D,E lần lượt là trung điểm của AB,AC

=>DE là đường trung bình của ΔABC

=>DE//BC và \(DE=\dfrac{BC}{2}\)

=>DE//HM

ΔHAC vuông tại H

mà HE là đường trung tuyến

nên \(HE=AE\)

mà AE=DM(cmt)

nên HE=DM

Xét tứ giác DHME có DE//HM

nên DHME là hình thang

Hình thang DHME có DM=HE

nên DHME là hình thang cân

\(8x^3+(1-3x)^3+(x-1)^3\\=8x^3+[1^3-3\cdot1^2\cdot3x+3\cdot1\cdot(3x)^2-(3x)^3]+(x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3)\\=8x^3+(1-9x+27x^2-27x^3)+(x^3-3x^2+3x-1)\\=8x^3+1-9x+27x^2-27x^3+x^3-3x^2+3x-1\\=(8x^3-27x^3+x^3)+(27x^2-3x^2)+(-9x+3x)+(1-1)\\=-18x^3+24x^2-6x\)

Đề bài là gì thế bạn?