Question

помогите пж
(3x-1/3y^2)² ​

Answer 1

Ответ.

$\Big(3x-\dfrac{1}{3y^2}\Big)^2=9x^2-2\cdot 3x\cdot \dfrac{1}{3y^2}+\dfrac{1}{9y^4}=9x^2-\dfrac{6x}{3y^2}+\dfrac{1}{9y^4}$

$\star \ \ \Big(\dfrac{3x-1}{3y^2}\Big)^2=\dfrac{(3x-1)^2}{(3y^2)^2}=\dfrac{9x^2-6x+1}{9y^4}\ \ \star$

Answer 2

Ответ:

$(3x-\dfrac{1}{3} y^{2} )^{2} =(3x)^{2}-2*3x*\dfrac{1}{3}y^{2}+(\dfrac{1}{3}y^{2} )^{2} = 9x^{2} -2xy^{2} +\dfrac{1}{9} y^{4} .$

$(3x-\dfrac{1}{3y^{2} } )^{2} =9x^{2} -2*3x*\dfrac{1}{3y^{2} } +\dfrac{1}{9y^{4} } =9x^{2} -2x/y^{2} +\dfrac{1}{9y^{4} } .$

Объяснение:

$(a-b)^{2}=a^{2} -2ab+b^{2} .$

$(a^{m} )^{n} =a^{mn} .$