Question

Помогите пожалуйста,с матерью ругаемся, решить не можем

Answer 1

$\boxed{\; A^2-B^2=(A-B)(A+B)\; }\\\\\\1)\; \; (x-5)(5+x)=(x-5)(x+5)=x^2-5^2=x^2-25\\\\2)\; \; (8+y)(y-8)=(y-8)(y+8)=y^2-8^2=y^2-64\\\\3)\; \; (10-k)(k+10)=10^2-k^2=100-k^2\\\\4)\; \; (a+\frac{2}{3}b)(a-\frac{2}{3}b)=a^2-(\frac{2}{3}b)^2=a^2-\frac{4}{9}b^2\\\\5)\; \; (\frac{4}{9}x-y)(y+\frac{4}{9}x)=(\frac{4}{9}x)^2-y^2=\frac{16}{81}x^2-y^2$

$6)\; \; (\frac{4}{15}m-n)(m+ \frac{4}{15}m)=\frac{16}{225}m^2-n^2\\\\7)\; \; (9x-5y)(9x+5y)=81x^2-25y^2\\\\8)\; \; (-4a+3b)(3b+4a)=(3b-4a)(3b+4a)=9b^2-16a^2\\\\9)\; \; (13k-2d)(2d+13k)=169k^2-4d^2\\\\10)\; \; (\frac{5}{4}c+\frac{3}{7}d)(\frac{3}{7}d-\frac{5}{4}c)=\frac{9}{49}d^2-\frac{25}{16}c^2\\\\11)\; \; (\frac{1}{3}x-3y)(3y+\frac{1}{3}x)=\frac{1}{9}x^2-9y^2\\\\12)\; \; (\frac{1}{5}a+\frac{1}{9}b)(\frac{1}{9}b-\frac{1}{5}a)=\frac{1}{81}b^2-\frac{1}{25}a^2$