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Question 1

Question 2

1.

$\left \{ {{\frac{x^2}{25}+\frac{y^2}{9}=1,} \atop {x+3y=3;}} \right.\left \{ {{x=3-3y,} \atop {\frac{(3-3y)^2}{25}+\frac{y^2}{9}-1=0;}} \right. \\ 81-162y+81y^2+25y^2-225=0, \\ 106y^2-162y-144=0, \\ 53y^2-81y-72=0, \\ D=21825, \\ y_1=\frac{81-15\sqrt{97}}{106}, y_2=\frac{81+15\sqrt{97}}{106}, \\ x_1=3-3\cdot\frac{81-15\sqrt{97}}{106}=\frac{75+45\sqrt{97}}{106}, \\ x_2=\frac{75-45\sqrt{97}}{106}. \\$

$(\frac{75+45\sqrt{97}}{106};\frac{81-15\sqrt{97}}{106}), (\frac{75-45\sqrt{97}}{106};\frac{81+15\sqrt{97}}{106}) \\$

2.

$y'=(2\sin x(1-\cos x))'=2(\sin x(1-\cos x))'= \\ =2((\sin x)'(1-\cos x)+\sin x(1-\cos x)')= \\ =2(\cos x(1-\cos x)+\sin x(1'-(\cos x)'))= \\ =2(\cos x-\cos^2x+\sin x(0-(-\sin x)))= \\ =2(\cos x-\cos^2x+\sin^2 x)=2(\cos x-\cos2x)=2\cos x-2\cos2x, \\$

$y'=(\cos x\cdot\sin^3x)'=(\cos x)'\cdot\sin^3x+\cos x\cdot(\sin^3x)'= \\=-\sin x\cdot\sin^3x+\cos x\cdot3\sin^2x\cdot(\sin x)'= \\ =-\sin^4x+3\cos x\sin^2x\cdot\cos x=-\sin^4x+3\cos^2x\sin^2x, \\$

$y'=(\ln\sqrt{2-3x}+\frac{3}{2-3x})'=(\ln\sqrt{2-3x})'+(\frac{3}{2-3x})'= \\ =\frac{1}{\sqrt{2-3x}}\cdot(\sqrt{2-3x})'+3(-\frac{(2-3x)'}{(2-3x)^2})= \\ =\frac{1}{\sqrt{2-3x}}\cdot\frac{1}{2\sqrt{2-3x}}\cdot(2-3x)'+3(-\frac{-3}{(2-3x)^2})= \\ =\frac{1}{2(\sqrt{2-3x})^2}\cdot(-3)+\frac{9}{(2-3x)^2}=-\frac{3}{2(2-3x)}+\frac{9}{(2-3x)^2}; \\$

3.

$v_1(t)=s_1'(t)=(2t^3-5t^2-3t)'=6t^2-10t-3, \\ v_2(t)=s_2'(t)=(2t^3-3t^2-11t+7)'=6t^2-6t-11, \\ v_1(t)=v_2(t), \\ 6t^2-10t-3=6t^2-6t-11, \\ 4t=7, \\ t=1,75, \\ a_1(t)=s_1''(t)=(6t^2-10t-3)'=12t-10, \\ a_1(\frac{7}{4})=11, \\ a_2(t)=s_2''(t)=(6t^2-6t-11)'=12t-6, \\ a_2(\frac{7}{4})=15. \\$