`(ax+by)^2-(ay+bx)^2`

`=(ax+by+ay+bx)(ax+by-ay-bx)`

`=[a(x+y)+b(x+y)][a(x-y)-b(x-y)]`

`=(x+y)(a+b)(x-y)(a-b)`

Bổ sung:

\(=\left(y+z\right)\left[16x^2\left(z-y\right)-\left(z-y\right)\right]\\ =\left(y+z\right)\left(z-y\right)\left(16x^2-1\right)\\ =\left(y+z\right)\left(z-y\right)\left(4x-1\right)\left(4x+1\right)\)

`(x^2-25)/(5x-x^2)=((x+5)(x-5))/(x(5-x))=(-(x+5)(x-5))/(x(x-5))=(-x-5)/x`

`16x^2z^2+y^2-z^2-16x^2y^2`

`=16x^2(z^2-y^2)+(y^2-z^2)`

`=16x^2(z-y)(y+z)+(y-z)(y+z)`

`=(y+z)[16x^2(z-y)+y-z]`

`=(y+z)(16x^2z-16x^2y+y-z)`

`x^3+4x^2+4x+3=0`

`<=>x^3+(x^2+3x^2)+(x+3x)+(1+3)=0`

`<=>(x^3+x^2+x)+(3x^2+3x+3)=0`

`<=>x(x^2+x+1)+3(x^2+x+1)=0`

`<=>(x+3)(x^2+x+1)=0`

`<=>x+3=0\ (x^2+x+1>0\ forall x)`

`<=>x=-3`

a, `(x-9)^4=(x-9)^7`

`(x-9)^4-(x-9)^7=0`

`(x-9)^4 . [(1-(x-9)^3]=0`

TH1: `(x-9)^4=0`

`x-9=0`

`x=9`

TH2: `1-(x-9)^3=0`

`(x-9)^3=1^3`

`x-9=1`

`x=10`

b, `(3x-15)^10=(3x-15)^15`

`(3x-15)^10 . [1-(3x-15)^5]=0`

TH1: `(3x-15)^10=0`

`3x-15=0`

`x=5`

TH2: `1-(3x-15)^5=0`

`(3x-15)^5=1^5`

`3x-15=1`

`x=16/3` (Loại)

c, `(x-8)^3=(x-8)^6`

`(x-8)^3 .[1-(x-8)^3]=0`

TH1: `(x-8)^3=0`

`x=8`

TH2: `1-(x-8)^3=0`

`x-8=1`

`x=9`

`\sqrt(4-\sqrt7). \sqrt(4+\sqrt7) =\sqrt((4-\sqrt7)(4+\sqrt7))=\sqrt(4^2-\sqrt7^2)=\sqrt(16-7)=\sqrt9=3`

`54^13 :2^13 : 9^13 : 3^12 = (2.3^3)^13 : 2^13 : (3^2)^13 :3^12`

`=3^39 .2^13 : 2^13 : 3^26 :3^12 = 3^39 :3^26 : 3^12 =3^13 : 3^12 =3`