Нужно с высшей с решениями дайте ответ! только 5 вариант. 4.5 6.5 7.5 15 числа сессия

Ответы

ElvirO 28.05.2020 20:49

$4)\; \; A_1(4,4,10)\; ,\; A_2(7,10,2)\; ,\; A_3(2,8,4)\; ,\; A_4(9,6,9)\\\\\overline {A_1A_2}=(3,6,-8)\; ,\; \overline {A_1A_3}=(-2,4,-6)\; ,\; \overline {A_1A_4}=(5,2,-1)\\\\(\overline {A_1A_2}\; ,\; \overline {A_1A_3}\, \, ,\, \overline {A_1A_4})=\left|\begin{array}{ccc}3&6&-8\\-2&4&-6\\5&2&-1\end{array}\right|=\\\\=3(-4+12)-6(2+30)-8(-4-20)=24-192+192=24\\\\V_{piramidu}=\frac{1}{6}\cdot 24=4\\\\V_{piramidu} =\frac{1}{3}\cdot h\cdot S_{osnovanija}=4\; \; \to \; \; h=\frac{12}{S_{osn.}}$

$\overline {A_1A_2}\times \overline {A_1A_3}=\left|\begin{array}{ccc}i&j&k\\3&6&-8\\-2&4&-6\end{array}\right|=\\\\=i(-36+32)-j(-18-16)+k(12+12)=-4i+34j+24k\\\\S_{A_1A_2A_3}=\frac{1}{2}\cdot \sqrt{4^2+34^2+24^2}=\frac{1}{2}\sqrt{1748}=\sqrt{437}\\\\h=\frac{12}{\sqrt{437}}$

$6)\; a)\; \; y=7x+\frac{5}{x^2}+\frac{6}{x}-\sqrt[7]{x^4}\; \; ,\; \; \; \; (\sqrt[7]{x^4}=x^{\frac{4}{7}})\\\\y'=7+5\cdot (-2)x^{-1}+6\cdot \frac{-1}{x^2}-\frac{4}{7}\cdot x^{-\frac{3}{7}}=7-\frac{10}{x}-\frac{6}{x^2}-\frac{4}{7\sqrt[7]{x^3}}\\\\b)\; \; y=\sqrt[5]{7x^2-x+5}-\frac{3}{(x-5)^4}\\\\y'=\frac{1}{5}\cdot (7x^2-x+5)^{-\frac{4}{5}}\cdot (14x-1)-\frac{-3\cdot 4(x-5)^3}{(x-5)^8}=\\\\=\frac{1}{5}\sqrt[5]{(7x^2-x+5)^4}+\frac{12}{(x-5)^5} \\\\c)\; \; y=ctg3x\cdot arccos(3x^2)$

$y'=-\frac{1}{sin^23x}\cdot 3\cdot arccos (3x^2)+ctg3x\cdot \frac{-1}{\sqrt{1-(3x^2)^2}}\cdot 6x=\\\\=-\frac{3\cdot arccos(3x^2)}{sin^23x}-\frac{6x\cdot ctg3x}{\sqrt{1-9x^4}}$

$7)\; \; z=x^2+xy+y^2+x-y+5\\\\z'_{x}=2x+y+1=0\; ,\; \; z'_{y}=x+2y-1=0\; \; ,\; \; \left \{ {{2x+y=-1} \atop {x+2y=1\, |\cdot (-2)}} \right. \oplus \\\\\left \{ {{2x+y=-1} \atop {-3y=-3}} \right. \; \left \{ {{x=-1} \atop {y=1}} \right. \; \; \Rightarrow \; \; \; M(-1,1)\\\\z''_{xx}=2\; \; ,\; \; z''_{yy}=2\; \; ,\; \; z''_{xy}=1\\\\A=z''_{xx}(M)=2\; \; ,\; \; B=z'_{yy}(M)=2\; \; ,\; \; C=z''_{xy}(M)=1\\\\\Delta =AB-C^2=2\cdot 2-1=30\; \; \; \to \; \; est'\; \; extremum\\\\A=z''_{xx}(M)=20\; \; \; \to \; \; min\\\\z_{min}=z(M)=z(-1,1)=(-1)^2-1\cdot 1+1^2-1-1+5=4$