Задачи от 2-ра степен с 2 неизвестни.

[SouthWinD]

43, 45, 47, 51, 53, и останалите. Моля ви решете ги защото му трябват ми за утре.
http://prikachi.com/images/872/4296872I.jpg
http://prikachi.com/images/877/4296877H.jpg
http://prikachi.com/images/879/4296879k.jpg
http://prikachi.com/images/881/4296881q.jpg

[ammornil]

не мога да заредя никоя от снимките!

[SouthWinD]

Извинявай нещо сайта не работи така както трябва
http://imageshack.us/photo/my-images/14/dsc00370um.jpg/
http://imageshack.us/photo/my-images/19 ... 373st.jpg/
http://imageshack.us/photo/my-images/59 ... 374rq.jpg/

[mail_dinko]

Задача 43.
[tex]\begin{tabular}{|l}x^2+y^2+4x=1\\x+y=-1=0--->\fbox{y=-1-x} \end{tabular}[/tex]
Заместваме в първото
[tex]x^2+y^2+4x=1[/tex]
[tex]x^2+(-1-x)^2+4x=1[/tex]
[tex]x^2+[-(1+x)]^2+4x=1[/tex]
[tex]x^2+(1+x)^2+4x-1=0[/tex]
[tex]x^2\cancel{+1}+2x+x^2+4x\cancel{-1}=0[/tex]
[tex]2x^2+6x=0|:2[/tex]
[tex]x^2+3x=0[/tex]
[tex]x(x+3)=0[/tex]
[tex]x_1=0--->y_1=-1-x=-1-0=-1--->(0;-1)[/tex]
[tex]x_2+3=0--->x_2=-3--->y_2=-1-x=-1-(-3)=-1+3=2--->(-3;2)[/tex]

Задача 45.
[tex]\begin{tabular}{|l}2x^2+2xy+y^2+4x-21=0\\2x+y-5=0 --->\fbox{y=5-2x}\end{tabular}[/tex]

[tex]2x^2+2x(5-2x)+(5-2x)^2+4x-21=0[/tex]
[tex]2x^2+10x\cancel{-4x^2}+25-20x\cancel{+4x^2}+4x-21=0[/tex]
[tex]2x^2-6x+4=0|:2[/tex]
[tex]x^2-3x+2=0[/tex]
[tex]D=9-8=1=1^2[/tex]
[tex]x_1=\frac{3+1}{2}=2--->y_1=5-2x=5-2.2=5-4=1--->(2;1)[/tex]
[tex]x_2=\frac{3-1}{2}=1-->y_2=5-2x=5-2.1=5-2=3--->(1;3)[/tex]

Задача 47.
[tex]\begin{tabular}{|l}4x^2+5y^2-5xy-2=0\\3x-y-2 =0--->\fbox{y=3x-2}\end{tabular}[/tex]
[tex]4x^2+5(3x-2)^2-5x(3x-2)-2=0[/tex]
[tex]4x^2+5(9x^2-12x+4)-15x^2+10x-2=0[/tex]
[tex]4x^2+45x^2-60x+20-15x^2+10x-2=0[/tex]
[tex]34x^2-50x+18=0|:2[/tex]
[tex]17x^2-25x+9=0[/tex]
[tex]D=625-612=13=(\sqrt{13})^2[/tex]
[tex]x_1=\frac{25+\sqrt{13}}{2.17}=\frac{25+\sqrt{13}}{34}--->y_1=3x-2=\frac{3.(25+\sqrt{13)}}{34}-\frac{68}{34}=\frac{75+3\sqrt{13}-68}{34}=\frac{7+3\sqrt{13}}{34}--->[/tex]
[tex]--->(\frac{25+\sqrt{13}}{34};\frac{7+3\sqrt{13}}{34})[/tex]
[tex]x_2=\frac{25-\sqrt{13}}{2.17}=\frac{25-\sqrt{13}}{34}--->y_2=3x-2=\frac{3.(25-\sqrt{13)}}{34}-\frac{68}{34}=\frac{75-3\sqrt{13}-68}{34}=\frac{7-3\sqrt{13}}{34}--->[/tex]
[tex]--->(\frac{25-\sqrt{13}}{34};\frac{7-3\sqrt{13}}{34})[/tex]

[mail_dinko]

Задача 51.
[tex]\begin{tabular}{|l}\frac{2x+y}{x-y}=1\\x^2+y^2+3x-2y=4\end{tabular}[/tex]

[tex]\begin{tabular}{|l}2x+y=x-y\\x^2+y^2+3x-2y=4\end{tabular}[/tex]

[tex]\begin{tabular}{|l}2x+y-x+y=0\\x^2+y^2+3x-2y=4\end{tabular}[/tex]

[tex]\begin{tabular}{|l}x+2y=0--->\fbox{x=-2y}\\x^2+y^2+3x-2y=4\end{tabular}[/tex]
[tex]x^2+y^2+3x-2y=4[/tex]
[tex](-2y)^2+y^2+3(-2y)-2y=4[/tex]
[tex]4y^2+y^2-6y-2y-4=0[/tex]
[tex]5y^2-8y-4=0[/tex]
[tex]D=16+20=36=6^2[/tex]
[tex]y_1=\frac{4+6}{5}=2--->x_1=-2y=-2.2=-4--->(-4;2)[/tex]
[tex]y_2=\frac{4-6}{5}=-\frac{2}{5}--->x_2=-2.(-\frac{2}{5})=\frac45--->(\frac45;-\frac25)[/tex]

Задача 53.
[tex]\begin{tabular}{|l}(2x-1)(x+y-3)=9\\\frac{2x-1}{x+y-3}=1\end{tabular}[/tex]

[tex]\begin{tabular}{|l}(2x-1)(x+y-3)=9\\\ 2x-1=x+y-3\end{tabular}[/tex]

[tex]\begin{tabular}{|l}(2x-1)(x+y-3)=9\\\ 2x-1-x-y+3=0\end{tabular}[/tex]

[tex]\begin{tabular}{|l}(2x-1)(x+y-3)=9\\\ x-y+2=0--->\fbox{x=y-2}\end{tabular}[/tex]
[tex](2x-1)(x+y-3)=9[/tex]
[tex](2(y-2)-1)(y-2+y-3)=9[/tex]
[tex](2y-4-1)(2y-5)=9[/tex]
[tex](2y-5)(2y-5)=9[/tex]
[tex](2y-5)^2=9[/tex]
[tex](2y-5)^2=(\pm3)^2[/tex]
[tex]2y-5=3--->2y=8|:2--->y_1=4--->x_1=y-2=4-2=2--->(2;4)[/tex]
[tex]2y-5=-3--->2y=2|:2--->y_2=1--->x_2=y-2=1-2=-1--->(-1;1)[/tex]

[mail_dinko]

Задача, на която не и се вижда номерът
[tex]\begin{tabular}{|l}x^2-2xy+x=-9\\2y-3x=1 \end{tabular}--->\fbox{y=\frac{1+3x}{2}}[/tex]
[tex]x^2-2xy+x=-9[/tex]
[tex]x^2-\cancel{2}x.\frac{1+3x}{\cancel{2}}+x+9=0[/tex]
[tex]x^2-x(1+3x)+x+9=0[/tex]
[tex]x^2\cancel{-x}-3x^2\cancel{+x}+9=0[/tex]
[tex]-2x^2+9=0[/tex]
[tex]-2x^2=-9|:(-2)[/tex]
[tex]x^2=\frac{-9}{-2}=\frac92[/tex]
[tex]x_{1,2}= \pm \sqrt {\frac92}= \pm \frac{3}{\sqrt{2}}.\frac{\sqrt{2}}{\sqrt{2}}=\pm \frac{3\sqrt{2}}{2}[/tex]
[tex]x_1= \frac{3\sqrt{2}}{2}--->y_1=\frac{1+3.\frac{3\sqrt{2}}{2}}{2}=\frac{\frac22+\frac{9\sqrt{2}}{2}}{2}=\frac{\frac{2+9\sqrt{2}}{2}}{2}=\frac{2+9\sqrt{2}}{4}--->(\frac{3\sqrt{2}}{2};\frac{2+9\sqrt{2}}{4})[/tex]
[tex]x_2=- \frac{3\sqrt{2}}{2}--->y_2=\frac{1-3.\frac{3\sqrt{2}}{2}}{2}=\frac{\frac22-\frac{9\sqrt{2}}{2}}{2}=\frac{\frac{2-9\sqrt{2}}{2}}{2}=\frac{2-9\sqrt{2}}{4}--->(-\frac{3\sqrt{2}}{2};\frac{2-9\sqrt{2}}{4})[/tex]

[SouthWinD]

Спасихте ми живота ! БЛАГОДАРЯ !

[mail_dinko]

чак пък живота, , тези задачи не кой знае какво- ако няма "дупки" по математика, би трябвало да можеш да ги решаваш