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

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

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

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

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

$Toru$

Lời giải:
Ta có:

$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=2^2-2(-23)=4+46=50$

Lời giải:

$M=x^3+y^3+2xy=(x+y)(x^2-xy+y^2)+2xy=x^2-xy+y^2+2xy$

$=x^2+y^2+xy=\frac{1}{4}(x-y)^2+\frac{3}{4}(x+y)^2=\frac{1}{4}(x-y)^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$. Giá trị này đạt được khi $x=y=\frac{1}{2}$

bbbbbbbbbbbbffv

Đề là $x(x+3)^3$ hay $x(x+3)^2$ hả bạn?

Lời giải:

a. Xét tam giác $AME$ và $AHE$ có:

$AE$ chung

$\widehat{AEM}=\widehat{AEH}=90^0$

$ME=HE$ (gt)

$\Rightarrow \triangle AME=\triangle AHE$(c.g.c)

$\Rightarrow AM=AH(1)$

Hoàn toàn tương tự ta có $\triangle AHF=\triangle ANF$ (c.g.c)

$\Rightarrow AH=AN(2)$

Từ $(1); (2)\Rightarrow AM=AN$ nên tam giác $AMN$ là tam giác cân tại $A$.

b.

Ta có:

$\frac{HE}{EM}=\frac{HF}{FN}=1$ nên theo định lý Talet thì $EF\parallel MN$ 

c.

Vì tam giác $AMN$ cân tại $A$ (cm ở phần a) nên trung tuyến $AI$ đồng thời là đường cao.

$\Rightarrow AI\perp MN$

Mà $MN\parallel EF$

$\Rightarrow AI\perp EF$ (đpcm)

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