![\displaystyle \lim_{n \to \infty}\dfrac{\sqrt{n^3+3}-\sqrt{n-3}}{\sqrt[5]{n^5+3}+\sqrt{n-3}}=\lim_{n \to \infty}\dfrac{n\sqrt{n}\cdot \sqrt{1+\dfrac{3}{n^3}}-\sqrt{n}\sqrt{1-\dfrac{3}{n}}}{n\sqrt[5]{1+\dfrac{3}{n^5}}+\sqrt{n}\sqrt{1-\dfrac{3}{n}}}\\ \\ \\ \\ =\lim_{n \to \infty}\dfrac{n\sqrt{n}\left(\sqrt{1+\dfrac{3}{n^3}}-\dfrac{1}{n}\sqrt{1-\dfrac{3}{n}}\right)}{n\left(\sqrt[5]{1+\dfrac{3}{n^5}}+\dfrac{1}{\sqrt{n}}\sqrt{1-\dfrac{3}{n}\right)}}=\lim_{n \to \infty}\dfrac{\sqrt{n}\cdot 1}{1}=\infty](/tpl/images/0982/6776/a1a51.png)
1) -3х+6у-12х-9у= -15x-3y
2) 6mn-2m-11mn-3n-5m=-5mn-7m-3n
1) (3a-7b)-(4a+8b)= 3a-7b-4a-8b=-a-15b
2)-(5m-7n)+(2n+12m)=-5m+7n+2n+12m=7m+9n
3) 3x(1-4x)-5x(6x+7) =3x-12x-30x-35x=-74x
4) 5c(2c+a)+(3c-2a)(5a-2c)=10c^2+5ca+15ca+6c^2-10a^2+4ca=16c^2+24ca-10a^2
5) (5y-3) куб. -(2-5y)куб=125y^3-225y^2+45y-27-8+150y - 60y^2+125y^3 =250y^3-285y^2+195y-32
1) 13(а-2)+10(4-а)=23
13a-26+40-10a=23
3a=9
a=3
2) (2х-1)(х+1)-х куб.=(х-3)куб -10
2x^2+2x-x-1-x^3=x^3-6x^2+27x-10
8x^2-28x-2x^3=-9
x(8x-28-2x^2)=-9
x1=0 (8x-28-2x^2)=-9
-2x^2+8x-19=0
D=8^2-4*(-2)-(-19)=-88(нет корней)
ответ:0
3) x/4 + x/8 =3/2
3x/8=3/2
3x=8*3/2
3x=12
x=4
