Question

Номер 1,2,3 Даю 33 б.

Answer 1

1)

$1) 4x^2-81y^2=(2x-9y)(2x+9y)\2) a^2b^2-frac{16}{9} =(ab-frac{4}{3} )(ab+frac{4}{3} )\3) 1,69y^{14}-900z^8=(1,3y^7-30z^4)(1,3y^7+30z^4)\4) -1+36a^6b^{10}=(6a^3b^5-1)(6a^3b^5+1)\5) 1frac{24}{25} m^6n^4-1frac{9}{16} a^2b^8=(frac{7}{5} m^3n^2-frac{5}{4} ab^4)(frac{7}{5} m^3n^2+frac{5}{4} ab^4)\6) 3^{4k}-25= ( 3 k - 2 ) * ( 3 k + 2 )$

2)

$1) (x-3)^2-25=(x-3-25)(x+3-25)=(x-28)(x-22)\2) a^6-(a+4)^2=(a^3-a-4)(a^3+a+4)$

3)

$(3x+4)^2-(2x-3)^2=0\(3x+4-2x+3)(3x+4+2x-3)=0\(x+7)(5x+1)=0\x_1=-7\x_2=-frac{1}{5}$