a: =x^3(x-y)+(x-y)

=(x-y)(x^3+1)

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

b: =(a-1)^2-9b^2

=(a-1-3b)(a-1+3b)

d) (8a3 – 27b3) – 2a(4a2 – 9b2)

= (2a – 3b)(4a2 + 6ab + 9b2) – 2a(2a – 3b)(2a + 3b)

= (2a – 3b)(4a2 + 6ab + 9b2 – 4a2 – 6ab) = 9b2(2a – 3b)

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

a)3y(x² -2xy+y)

b)=(a+b)(a-b)+2(a+b)

=(a+b)(a-b+2)

b)

`a)x^4+2x^2y+y^2`

`=(x^2+y)^2`

`b)(2a+b)^2-(2b+a)^2`

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

`=(a-b)(3a+3b)`

`=3(a-b)(a+b)`

`c)8a^3-27b^3-2a(4a^2-9b^2)`

`=(2a-3b)(4a^2+6ab+9b^2)-2a(2a-3b)(2a+3b)`

`=(2a-3b)(4a^2+6ab+9b^2-3a^2-6ab)`

`=9b^2(2a-3b)`

a) Ta có: \(x^4+2x^2y+y^2\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot y+y^2\)

\(=\left(x^2+y\right)^2\)

b) Ta có: \(\left(2a+b\right)^2-\left(2b+a\right)^2\)

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

\(=\left(a-b\right)\left(3a+3b\right)\)

\(=3\left(a+b\right)\left(a-b\right)\)

a, \(4abc-8ab^2c=4abc\left(1-2b\right)\)

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

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

c, \(9a^4\left(a-2\right)+a^2\left(a-2\right)=a^2\left(9a^2+1\right)\left(a-2\right)\)

d, \(\left(a-4\right)\left(2a-1\right)-8a+4=\left(a-4\right)\left(2a-1\right)-4\left(2a-1\right)\)

\(=\left(a-8\right)\left(2a-1\right)\)

a) `4abc-8ab^2c=4abc(1-2b)`

b) `x^2 (2a-1)+x(1-2a) = x^2 (2a-1) -x(2a-1) = (2a-1)(x^2-x)=x(2a-1)(x-1)`

c) `9a^4 (a-2) +a^2 (a-2) = (a-2)(9a^4+a^2)=a^2 (a-2)(9a^2+1)`

d) `(a-4)(2a-1)-8a+4=(a-4)(2a-1)-4(2a-1)=(2a-1)(a-8)`

\(a,Sửa:a^2-b^2=\left(a-b\right)\left(a+b\right)\\ b,=a^4+2a^2b^2+b^4-2a^2b^2\\ =\left(a^2+b^2\right)^2-2a^2b^2=\left(a^2+b^2-ab\sqrt{2}\right)\left(a^2+b^2+ab\sqrt{2}\right)\\ c,=a\left(a-1\right)\\ d,=a^2-a-2a+2=\left(a-1\right)\left(a-2\right)\\ e,=a^2-2a-3a+6=\left(a-2\right)\left(a-3\right)\\ g,=a^2-3a-4a+12=\left(a-3\right)\left(a-4\right)\)

a) \(a^2+ab-7a-7b=a\left(a+b\right)-7\left(a+b\right)=\left(a+b\right)\left(a-7\right)\)

b) \(5ab+4c+20b+ac=5b\left(a+4\right)+c\left(a+4\right)=\left(a+4\right)\left(5b+c\right)\)

c) \(a^2+6a-b^2+9=\left(a+3\right)^2-b^2=\left(a+b-b\right)\left(a+3+b\right)\)

d) \(a^2-16=\left(a-4\right)\left(a+4\right)\)

TK
 

a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).

b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2

= (a + 4 – b)(a + 4 + b).