Question

Выполните действие: (х-5)(5+х); (10-k)(k+10); (49x-y)(y+49x); (9x-5y)(9x+5y)

Answer 1

Объяснение:

1)

$(x-5)(5+x)= x^{2} -5^{2} =x^{2} -25;$

3)

$(10-k) (k+10) = 10^{2} -k^{2} =100- k^{2} ;$

5)

$(frac{4}{9} x-y) (y+frac{4}{9} x) = (frac{4}{9} x)^{2} -y^{2} = frac{16}{81} x^{2} -y^{2} ;$

7)

$(9x-5y) (9x+5y)=(9x)^{2} -(5y)^{2} =81x^{2} -25y^{2} .$

Answer 2

$1)(x-5)(5+x)=(x-5)(x+5)=x^2-25\2)(10-k)(k+10)=(10-k)(10+k)=100-k^2\3)(frac{4}{9}x-y)(y+frac{4}{9}x )=(frac{4}{9}x-y)(frac{4}{9}x+y )=frac{16}{81}x^2-y^2\4)(9x-5y)(9x+5y)=81x^2-25y^2$