Помогите решить 1,2,3,7,5 Срочно. Заранее спасибо.))

4х^2-12=04х^2=12 (:4)х^2=3х=под корнем 3

1.4x²-12=04(x²-3)=0x²-3=0x²=3x₁= -√3x₂=√3Ответ: -√3; √3.2.[4x/(x-y)(x+y)] - [4(x-y)/(x+y)(x-y)]=(4x-4x+4y)/(x²-y²)=(4y)/(x²-y²)3.3x>12+11x3x-11x>12-8x>12x<12 : (-8)x< -1.55x-1<05x<1x<0.2{x< -1.5{x<0.2\\\\\\\\\\\-------- -1.5 ---------- 0.2 ------------\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\x< -1.5x∈(-∞; -1.5)5.3x²+2x-1=3x²+3x-x-1=3x(x+1)-(x+1)=(x+1)(3x-1)7.sin²α-2sinαcosα+cos²α+2sinαcosα=sin²α+cos²α=1