Question

2)(1 4/7 x⁵ -z²) (1 4/5 x⁵ +z²)
4)(m⁶-2 1/3 n⁵) (2 1/3 n⁵+m6)​

Answer 1

Ответ:

$\boxed{\ (a-b)(a+b)=a^2-b^2\ }\\\\\\2)\ \ \Big(1\dfrac{4}{7}\ x^5-z^2\Big)\Big(1\dfrac{4}{5}\, x^5-z^2\Big)=\Big(\dfrac{11}{7}\, x^5-z^2\Big)\Big(\dfrac{9}{5}\, x^5+z^2\Big)=\\\\\\=\dfrac{99}{35}\, x^{10}+\dfrac{11}{7}\, x^5z^2-\dfrac{9}{5}\, x^5z^2-z^4=\dfrac{99}{35}\, x^{10}-\dfrac{8}{35}\, x^5z^2-z^4\\\\\\\star \ \Big(1\dfrac{4}{7}\ x^5-z^2\Big)\Big(1\dfrac{4}{7}\, x^5-z^2\Big)=\Big(\dfrac{11}{7}\, x^5-z^2\Big)\Big(\dfrac{11}{7}\, x^5+z^2\Big)=\dfrac{121}{49}\, x^{10}-z^4\ \ \star$

$4)\ \ \Big(m^6-2\dfrac{1}{3}\ n^5\Big)\Big(2\dfrac{1}{3}\, n^5+m^6\Big)=\Big(m^6-\dfrac{7}{3}\, n^5\Big)\Big(m^6+\dfrac{7}{3}\, n^5\Big)=\\\\\\=m^{12}-\dfrac{49}{9}\, n^{10}=m^{12}-5\dfrac{14}{9}\, n^{10}$