Question

помогите пожалуйста ​

Answer 1

б)

$\dfrac{81}{16}x^6 - 9x^3y^2 + 4y^4 = \left(\dfrac{9}{4}x^3\right)^2 - 2\cdot\dfrac{9}{4}\cdot2\cdot x^3y^2 + (2y^2)^2 = \left(\dfrac{9}{4}x^3 - 2y^2\right)^2$

г)

$25 - (1-2a)^2 = 5^2 - (1-2a)^2 = (5-1+2a)(5+1-2a) = (4+2a)(6-2a) = 4(2+a)(3-a)$

Answer 2

Ответ:

$\bigg (\dfrac{9}{4}x^{3}-2y^{2} \bigg )^{2} \quad ; \quad (4+2a)(6-2a) \quad ;$

Объяснение:

$\dfrac{81}{16}x^{6}-9x^{3}y^{2}+4y^{4}=\bigg (\dfrac{9}{4}x^{3} \bigg )^{2}-2 \cdot \dfrac{9}{4}x^{3} \cdot 2y^{2}+(2y^{2})^{2}=\bigg (\dfrac{9}{4}x^{3}-2y^{2} \bigg )^{2};$

$25-(1-2a)^{2}=5^{2}-(1-2a)^{2}=(5-(1-2a)) \cdot (5+(1-2a))=(5-1+2a) \cdot$

$\cdot (5+1-2a)=(4+2a)(6-2a);$