Đáp án:

 Bài `1`

`a, (-6xy^5) .(3x^2 y)`

`= (-6 .3)(x.x^2).(y^5 .y)`

`= -18x^3 y^6`

`b, (4x^7 y^2)^2 .(-3xy^3)`

`= 16x^14 y^4 . (-3xy^3)`

`= [16 . (-3)](x^14 .x)(y^4 . y^3)`

`= -48x^15 y^7`

`c, 5x^4 - (2x).(3x^2) +(3x^3).(-4x) + 6x^3`

`= 5x^4 - 6x^3 - 12x^4 + 6x^3`

`= (5x^4 - 12x^4) +( -6x^3 + 6x^3)`

`= -7x^4`

Bài `2`:

`a, x(1/5)^9 = (1/5)^10`

`⇔ x = (1/5)^10 : (1/5)^9`

`⇔ x = (1/5)^(10 - 9)`

`⇔ x = (1/5)^1 = 1/5`

Vậy `x = 1/5`

`c, 3^(x - 12) : 81 = 27`

`⇔ 3^(x - 12) : 3^4 = 3^3`

`⇔ 3^(x - 12) = 3^3 . 3^4`

`⇔ 3^(x - 12) = 3^7`

`⇔ x - 12 = 7`

`⇒ x = 7 + 12`

`⇒ x = 19`

Vậy `x = 19`

`d, (5x - 1)^3 = -8`

`⇔ (5x - 1)^3 = (-2)^3`

`⇔ 5x - 1 = -2`

`⇔ 5x = -2 + 1`

`⇔ 5x = -1`

`⇔ x = -1/5`

Vậy `x = -1/5`

`e, (5 - 2x)^2 = 25`

`⇔ (5 - 2x)^2 - 25 = 0`

`⇔ (5 - 2x)^2 - 5^2 = 0`

`⇔ (5 - 2x + 5)(5 - 2x - 5) = 0`

`⇔ (10 - 2x).(-2x) = 0`

`⇒` $\left[\begin{matrix} -2x = 0\\ 10-2x = 0\end{matrix}\right.$

`⇒` $\left[\begin{matrix} x = 0\\ 2x = 10\end{matrix}\right.$

`⇒` $\left[\begin{matrix} x = 0\\ x = 5\end{matrix}\right.$

Vậy `x = 0` hoặc `x = 5`

`#Sad`

`1)`

`a)`

`(-6xy^5). (3x^2y)`

`= -18x^3y^6`

`b)`

`(4x^7y^2)^2. (-3xy^3)`

`= (16x^14y^4). (-3xy^3)`

`= -48x^(15)y^7`

`c)`

`5x^4-(2x). (3x^2)+(3x^3). (-4x)+6x^3`

`= 5x^4-6x^3-12x^4+6x^3`

`= (5x^4-12x^4)+(-6x^3+6x^3)`

`= -7x^4`

`2)`

`a)`

`x. (1/5)^9 = (1/5)^(10)`

`⇔ x = (1/5)^(10) : (1/5)^9`

`⇔ x = 1/5`

`\text{Vậy}` `x=1/5`

`c)`

`3^(x-12) : 81 = 27`

`⇔ 3^(x-12) = 2187`

`⇔ 3^(x-12) = 3^7`

`⇔ x-12 = 7`

`⇔ x = 19`

`\text{Vậy}` `x=19`

`d)`

`(5x-1)^3 = -8`

`⇔ (5x-1)^3 = (-2)^3`

`⇔ 5x-1 = -2`

`⇔ 5x = -1`

`⇔ x = -(1)/(5)`

`\text{Vậy}` `x=-(1)/(5)`

`e)`

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

`⇔ (5-2x)^2 = (+-5)^2`

`⇔` \(\left[ \begin{array}{l}5-2x=5\\5-2x=-5\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}2x=0\\2x=10\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=0\\x=5\end{array} \right.\) 

`\text{Vậy}` `x=0` `\text{hoặc}` `x=5`