Question

Разложить а)(7х-4)^2-(2х+1)^2 б) 2 1\4m^6 - 1 11\25 в) (5х-12)^2-х^2 г) 49(у-4)^2 - 9(у+2)^2 заранее огромное ))0)

Answer 1

А)
$(7x-4)^2-(2x+1)^2 =((7x-4)-(2x+1))((7x-4)+(2x+1)) = \\=(7x-4-2x-1)(7x-4+2x+1)=(5x-5)(9x-3)$
б)
$2\frac{1}{4}m^6 - 1\frac{11}{25}= \frac{9}{4}m^6 - \frac{36}{25}=( \frac{3}{2}m^3 - \frac{6}{5})(\frac{3}{2}m^3 + \frac{6}{5})= \\\\=(1,m^3-1,2)(1,5m^3+1,2)$
в)
 $(5x-12)^2-x^2=((5x-12)-x)((5x+12)-x)=\\=(5x-12-x)(5x-12+x)=(4x-12)(6x-12)$
г)
$49(y-4)^2 - 9(y+2)^2=(7(y-4) - 3(y+2))(7(y-4)+3(y+2))=\\=(7y-28 - 3y-6)(7y-28+3y+6)=(4y-34)(10y-22)$