a) 3x²y.(1/6x²y²z) b) -5x³y²+10x³y²+ (-3/4x³y²) - x³y²

Đáp án:

 `a, 1/2 x^4y^3z`

`b, 13/4 x^3y^2`

Giải thích các bước giải:

 `a, 3x^2y . (1/6x^2y^2z)`

` = (3 . 1/6)(x^2 . x^2)(y . y^2)z`

` = 1/2 x^4 y^3z`

`b, -5x^3y^2 + 10x^3y^2 + (-3/4x^3y^2) - x^3y^2`

` =  [ (-5) + 10 + (-3/4) - 1]x^3y^2 `

` = (5 - 7/4)x^3y^2`

` = (20/4 - 7/4)x^3y^2`

` = 13/4 x^3y^2`

$a)$ $3x²y\cdot \left(\dfrac{1}{6x²y²z}\right)$

$=3x^2y\dfrac{1}{6x^2y^2z}$

$=\dfrac{3x^2y\cdot \:1}{6x^2y^2z}$

$=\dfrac{3x^2y}{6x^2y^2z}$

$=\dfrac{3y}{6y^2z}$

$=\dfrac{3y}{3\cdot \:2y^2z}$

$=\dfrac{y}{2y^2z}$

$=\dfrac{y}{2y^2z}$

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$b)$ $-5x^3y^2+10x^3y^2+\left(\dfrac{-3}{4x^3y^2}\right)-x^3y^2$

$=-5x^3y^2+10x^3y^2+\dfrac{-3}{4x^3y^2}-x^3y^2$

$=4x^3y^2+\dfrac{-3}{4x^3y^2}$

$=4x^3y^2-\dfrac{3}{4x^3y^2}$

$=\dfrac{4x^3y^2\cdot \:4x^3y^2-3}{4x^3y^2}$

$=\dfrac{16x^6y^4-3}{4x^3y^2}$