a) 7x – 14 = 7(x - 2)

b) 5x3 - 10x2y +5xy2 = 5x(x² - 2xy - y²) = 5x(x - y)²

c) 25 – x2= (x - 5)(x + 5)

a, \(7\left(x-2\right)\)

c, \(\left(5-x\right)\left(5+x\right)\)

 \(a,7x-14=7\left(x-2\right)\\ b.5x^3-10x^2y+5xy^2=5x\left(x^2-2xy+y^2\right)=5x\left(x-y\right)^2\\ c,25-x^2=\left(5-x\right)\left(5+x\right)\)

a, 7x - 14 = 7 ( x - 2 )

b, 5x³ - 10x²y + 5xy² 

= 5x ( x² - 2xy + y² ) 

= 5x ( x - y )²

c, 25 - x² = 5² - x² = ( 5 - x ) ( 5 + x )

a) \(=3xy\left(2x-3\right)\)

b) \(=\left(y+5\right)^2\)

a. 7x⁵ - 14x³y + 21x²y

= 7x²(x³ - 2xy + 3y)

b. 4x² - 20x + 25

= (2x)² - 2x.2.5 + 5²

= (2x - 5)²

a) \(7x^5-14x^3y+21x^2y\)

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

b) \(4x^2-20x+25\)

\(=\left(2x\right)^2-2.2x.5+5^2\)

\(\left(2x-5\right)^2\)

a) \(3x-1=\left(\sqrt{3x}\right)^2-1^2=\left(\sqrt{3x}-1\right)\left(\sqrt{3x}+1\right)\)

b) \(4x-25=\left(2\sqrt{x}\right)^2-5^2=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)

c) \(x-3\sqrt{x}-4\left(x\ge0\right)\Rightarrow x+\sqrt{x}-4\sqrt{x}-4\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)\)

b) \(4x-25=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)

c) \(x-3\sqrt{x}-4=\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)\)

\(a,=\left(a-5\right)^2-4b^2=\left(a-2b-5\right)\left(a+2b-5\right)\\ b,=ax^2+a-a^2x-x=ax\left(a-x\right)+\left(a-x\right)=\left(ax+1\right)\left(a-x\right)\)

a: \(=\left(a-5-2b\right)\left(a-5+2b\right)\)

b: \(ax^2+a-a^2x-x\)

\(=ax\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(ax-1\right)\)

\(-9x^4+3x^2+2\\ =-9x^4+6x^2-3x^2+2\\ =-3x^2\left(3x^2-2\right)-\left(3x^2-2\right)\\ =-\left(3x^2-2\right)\left(3x^2+1\right)\)

\(-9x^4+3x^2+2\)

\(=-9x^4+6x^2-3x^2+2\)

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

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

a)10x-x²-25=-(x²-10x+25)=-(x-5)²

a) 10x-x²-25=10x-50-x²+25=(10x-50)-(x²-25)=10(x-5)-(x-5)(x+5)=(x-5)(5-x)

a) \(10x-x^2-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

Hửm đề sai rồi phải là:

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

`=a^2-ab-2ab+2b^2`

`=a(a-b)-2b(a-b)`

`=(a-b)(a-2b)`

a) \(25a^2-1=\left(5a-1\right)\left(5a+1\right)\)

b) \(a^2-9=\left(a-3\right)\left(a+3\right)\)

c) \(\dfrac{1}{4}a^2-\dfrac{9}{25}=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)\)

d) \(\dfrac{9}{4}a^4-\dfrac{16}{25}=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\)

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

f) \(16\left(x-1\right)^2-25\left(x+y\right)^2=\left(4x-4-5x-5y\right)\left(4x-4+5x+5y\right)=\left(-x-4-5y\right)\left(9x+5y-4\right)\)

a/ $25x^2-1\\=(5x)^2-1^2\\=(5x-1)(5x+1)$

b/ $a^2-9\\=a^2-3^2\\=(a-3)(a+3)$

c/ $\dfrac{1}{4}a^2-\dfrac{9}{25}\\=\left(\dfrac{1}{2}a\right)^2-\left(\dfrac{3}{5}\right)^2\\=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)$

d/ $\dfrac{9}{4}a^4-\dfrac{16}{25}\\=\left(\dfrac{3}{2}a^2\right)^2-\left(\dfrac{4}{5}\right)^2\\=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left[\left(\sqrt{\dfrac 3 2}a\right)^2-\left(\dfrac{2\sqrt 5}{5}\right)^2\right]\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left(\sqrt{\dfrac 3 2}a-\dfrac{2\sqrt 5}{5}\right)\left(\sqrt{\dfrac 3 2}a+\dfrac{2\sqrt 5}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)$

e/ $(2a+b)^2-a^2\\=(2a+b-a)(2a+b+a)\\=(a+b)(3a+b)$

f/ $16(x-1)^2-25(x+y)^2\\=[4(x-1)]^2-[5(x-y)]^2\\=[4(x-1)-5(x-y)][4(x-1)+5(x-y)]\\=[4x-4-5x+5y][4x-4+5x-5y]\\=(-x+5y-4)(9x-5y-4)$