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

4х^2-12=0 4х^2=12 (:4) х^2=3 х=под корнем 3

1. 4x²-12=0 4(x²-3)=0 x²-3=0 x²=3 x₁= -√3 x₂=√3 Ответ: -√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+11x 3x-11x>12 -8x>12 x<12 : (-8) x< -1.5 5x-1<0 5x<1 x<0.2 {x< -1.5 {x<0.2 \\\\\\\\\\\ -------- -1.5 ---------- 0.2 ------------ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ x< -1.5 x∈(-∞; -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