Question

Решите пожалуйста 100 баллов ​

Answer 1

$\displaystyle\bf\\1)\\\\3x^{2} -75=3(x^{2} -25)=3(x-5)(x+5)\\\\2)\\\\\frac{48^{2} -12^{2} }{89^{2} +2\cdot31\cdot89+31^{2} } =\frac{(48-12)(48+12)}{(89+31)^{2} } =\frac{36\cdot60}{120} =18\\\\3)\\\\y^{3} +\underbrace{(7-y)(49+7y+y^{2})}_{7^{3} -y^{3} } -1=y^{3}+7^{3} -y^{3} -1=343-1=342\\\\342=342$

$\displaystyle\bf\\4)\\\\(2x+3)^{2}-(2x-1)(2x+1)=46\\\\4x^{2} +12x+9-(4x^{2} -1)=46\\\\4x^{2} +12x+9-4x^{2} +1=46\\\\12x+10=46\\\\12x=36\\\\x=3$