\(x^5+12=13\\ \Rightarrow x^5=13-12\\ \Rightarrow x^5=1\\ \Rightarrow x^5=1^5\\ \Rightarrow x=1\)

Vậy `x=1` 

\(3\cdot4^7=3\cdot\left(2^2\right)^7=3\cdot2^{14}\)

\(8^5=\left(2^3\right)^5=2^{15}=2\cdot2^{14}\)

Mà `3>2`

Nên \(3\cdot4^7>8^5\)

`x^2 -6x+9-121y^2`

`=(x^2-6x+9)-(11y)^2`

`=(x-3)^2 -(11y)^2`

`=(x-3-11y)(x-3+11y)`

`4x^2 -81y^2`

`=(2x)^2 -(9y)^2`

`=(2x-9y)(2x+9y)`

V cho ra đảo ruii💞

-> iu ghiaa 👉👈🙆

`x^2 +4x+4-25y^2`

`=(x^2+4x+4) - (5y)^2`

`= (x+2)^2 -(5y)^2`

`= (x+2-5y)(x+2+5y)`

`8x^3 -27y^3`

`= (2x)^3 - (3y)^3`

`=(2x-3y)(4x^2 + 6xy+9y^2)`

`x^2+2xy+y^2-9`

`=(x^2+2xy+y^2) -3^2`

`=(x+y)^2 -3^2`

`=(x+y-3)(x+y+3)`

`x^3-125`

`=x^3-5^3`

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