Ответ:
Объяснение:
![(7-4\sqrt{3} )x^2+(2-\sqrt{3})x=2](https://tex.z-dn.net/?f=%287-4%5Csqrt%7B3%7D%20%29x%5E2%2B%282-%5Csqrt%7B3%7D%29x%3D2)
заметим, что:
7-4√3=2²-2*2*√3+(√3)²=(2-√3)²
(2-√3)²x²+(2-√3)x-2=0
D=b²-4ac
D=(2-√3)²+4*2(2-√3)²=(3*(2-√3))²
![\displaystyle\\x_1=\frac{-(2-\sqrt{3})-3(2-\sqrt{3})}{2*(2-\sqrt{3})^2}=\frac{-4(2-\sqrt{3})}{2(2-\sqrt{3})^2}=\frac{-2}{2-\sqrt{3}}*\frac{2+\sqrt{3} }{2+\sqrt{3}} =-4-2\sqrt{3}\\ \\ \\ x_2=\frac{-(2-\sqrt{3})+3(2-\sqrt{3})}{2*(2-\sqrt{3})^2}=\frac{2(2-\sqrt{3})}{2(2-\sqrt{3})^2}=\frac{1}{2-\sqrt{3}}*\frac{2+\sqrt{3} }{2+\sqrt{3}} =2+\sqrt{3}\\ \\ \\](https://tex.z-dn.net/?f=%5Cdisplaystyle%5C%5Cx_1%3D%5Cfrac%7B-%282-%5Csqrt%7B3%7D%29-3%282-%5Csqrt%7B3%7D%29%7D%7B2%2A%282-%5Csqrt%7B3%7D%29%5E2%7D%3D%5Cfrac%7B-4%282-%5Csqrt%7B3%7D%29%7D%7B2%282-%5Csqrt%7B3%7D%29%5E2%7D%3D%5Cfrac%7B-2%7D%7B2-%5Csqrt%7B3%7D%7D%2A%5Cfrac%7B2%2B%5Csqrt%7B3%7D%20%7D%7B2%2B%5Csqrt%7B3%7D%7D%20%3D-4-2%5Csqrt%7B3%7D%5C%5C%20%5C%5C%20%5C%5C%20x_2%3D%5Cfrac%7B-%282-%5Csqrt%7B3%7D%29%2B3%282-%5Csqrt%7B3%7D%29%7D%7B2%2A%282-%5Csqrt%7B3%7D%29%5E2%7D%3D%5Cfrac%7B2%282-%5Csqrt%7B3%7D%29%7D%7B2%282-%5Csqrt%7B3%7D%29%5E2%7D%3D%5Cfrac%7B1%7D%7B2-%5Csqrt%7B3%7D%7D%2A%5Cfrac%7B2%2B%5Csqrt%7B3%7D%20%7D%7B2%2B%5Csqrt%7B3%7D%7D%20%3D2%2B%5Csqrt%7B3%7D%5C%5C%20%5C%5C%20%5C%5C)
Ответ:y = C*exp(2*x) - 2
так как при х=0 у=-1 то
с-2=-1
с=1
и значит y = exp(2*x) - 2
Ответ ответ ответ ответ ответ ответ
2/3v7 = 2v7/28
4/(v11 +3) = 4(v11 - 3)/2= 2(v11 - 3)=2v11-6
А) b³-2b²+b=b(b²-2b+1)=b(b-1)²;
б)ab³+2a²b²+a³b=ab(b²+2ab+a²)=ab(a+b)²;
в)3a+3b-ax-bx=3(a+b)-x(a+b)=(a+b)·(3-x);
г)5a-b+5a²-ab=5a+5a²-b-ab=5a(1+a)-b(1+a)=(1+a)·(5a-b);
д)7a-7b+2b²-2ab=7(a-b)-2b(a-b)=(a-b)·(7-2b);
е)b⁴-b²+4b+4=b²(b²-1)+4(b+1)=b²(b-1)(b+1)+4(b+1)=
=(b+1)·(b²(b-1)+4)=(b+1)(b³-b²+4);