Lời giải:

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

b. $5x(x-2)-7(x-2)=(x-2)(5x-7)$

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

d. $4x(x-2)+5(2-x)=4x(x-2)-5(x-2)=(x-2)(4x-5)$

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

Ta có \(a^2+b^2+c^2\ge ab+bc+ca\)

<=> \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

P = \(\dfrac{a}{bc\left(a+c\right)}+\dfrac{b}{ac\left(a+b\right)}+\dfrac{c}{ab\left(b+c\right)}\)

\(=\dfrac{a^2}{abc\left(a+c\right)}+\dfrac{b^2}{abc\left(a+b\right)}+\dfrac{c^2}{abc\left(b+c\right)}\)

\(\ge\dfrac{\left(a+b+c\right)^2}{abc\left(a+c\right)+abc\left(a+b\right)+abc\left(b+c\right)}\) (BĐT Cauchy - Schwarz) 

\(=\dfrac{\left(a+b+c\right)^2}{2abc\left(a+b+c\right)}=\dfrac{a+b+c}{2abc}=\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)

\(\ge\dfrac{1}{2}.\dfrac{\left(1+1+1\right)^2}{ab+bc+ca}\) (BĐT Cauchy - Schwarz) 

\(=\dfrac{9}{2\left(ab+bc+ca\right)}\ge\dfrac{9}{\dfrac{2\left(a+b+c\right)^2}{3}}=\dfrac{27}{2\left(a+b+c\right)^2}\)

chịu ====

mình cx chịu

9(x + 5)^2 - (x + 7)^2
<=>[3(x+5)]^2 - (x+7)^2
<=>[(3(x+5)-(x+7)] [(3(x+5)+(x+7)]
<=>(3x+15-x-7)(3x+15+x+7)
<=>(2x+8)(4x+22)
<=>2(x+4) . 2(2x+11)
<=>4(x+4)(2x+11)

9(x + 5)^2 - (x + 7)^2
<=>[3(x+5)]^2 - (x+7)^2
<=>[(3(x+5)-(x+7)] [(3(x+5)+(x+7)]
<=>(3x+15-x-7)(3x+15+x+7)
<=>(2x+8)(4x+22)
<=>2(x+4) . 2(2x+11)
<=>4(x+4)(2x+11)

Bài 10:

Vì $b^2+c^2=2a^2\vdots 2$ nên $b,c$ cùng tính chẵn lẻ

Nếu $b,c$ cùng lẻ thì ta biết 1 số chính phương lẻ chia $8$ dư $1$ nên:

$2a^2\equiv 1+1\equiv 2\pmod 8$

$\Rightarrow a^2\equiv 1\pmod 8$

$\Rightarrow a$ lẻ 

Do đó: $a+b$ chẵn $(1)$

Nếu $a+b=2$ thì hiển nhiên $a=b=1\Rightarrow c=0$ (vô lý vì $c\in\mathbb{N}^*$)

$\Rightarrow a+b>2(2)$

Từ $(1); (2)\Rightarrow a+b$ là hợp số $(*)$

Nếu $b,c$ đều chẵn

$\Rightarrow 2a^2=b^2+c^2\equiv 0\pmod 4$

$\Rightarrow a^2\equiv 0\pmod 2$

$\Rightarrow a$ chẵn 

$\Rightarrow a+b$ chẵn. Hiển nhiên $a+b>2$ nên $a+b$ là hợp số $(**)$

Từ $(*); (**)$ ta có đpcm.

(2x+1)²-(x-1)²

= 4x²+4x+1-(x²-2x+1)

= 4x² + 4x + 1 - x² + 2x - 1

= 3x² + 6x = 3x(x+2)

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

1) \(\dfrac{1}{8}x^3-8=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2-x+4\right)\)

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

3) \(10x^3-10a=10a\left(a^2-1\right)=10a\left(a-1\right)\left(a+1\right)\)

4) \(2x^3+16=2\left(x^2+8\right)=2\left(x+2\right)\left(x^2-2x+4\right)\)

5) \(x^2-5x+6=\left(x^2-3x\right)-\left(2x-6\right)=x\left(x-3\right)-2\left(x-3\right)=\left(x-2\right)\left(x-3\right)\)

6) \(x^2+5x+6=\left(x^2+3x\right)+\left(2x+6\right)=x\left(x+3\right)+2\left(x+3\right)=\left(x+2\right)\left(x+3\right)\)

7) \(x^2-7x+12=\left(x^2-3x\right)-\left(4x-12\right)=x\left(x-3\right)-4\left(x-3\right)=\left(x-4\right)\left(x-3\right)\)

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

9) \(x^2+x-20=\left(x^2-4x\right)+\left(5x-20\right)=x\left(x-4\right)+5\left(x-4\right)=\left(x+5\right)\left(x-4\right)\)

chữ đẹp ghia nhìn là ko muốn trả lời =))