Question

Плиз решите #50,#51 ДАЮ 100 БАЛЛОВ

Answer 1

$\displaystyle \tt50.\\\\a). \ \ \left \{ {{x-y=1} \ \ \ \ \atop {x^{2}+2y=33}} \right.\\\\\\{} \ \ \ \ \ \left \{ {{x=y+1} \ \ \ \ \ \ \ \ \ \ \ \atop {(y+1)^{2}+2y=33}} \right.\\\\\\{} \ \ \ \ \ \ \ y^{2}+2y+1+2y=33\\\\{} \ \ \ \ \ \ \ y^{2}+4y-32=0 \ \ \ \ \ \ \ \ \ D=b^{2}-4ac=16+128=144\\\\{} \ \ \ \ \ \ \ y_{1,2}=\frac{-bб\sqrt{D}}{2a}=\frac{-4б\sqrt{144}}{2} \\\\{} \ \ \ \ \ \ \ y_{1}=-8 \ \ \ \ \ \ \ \ x_{1}=-8+1=-7 \\\\{} \ \ \ \ \ \ \ y_{2}=4 \ \ \ \ \ \ \ \ \ \ x_{2}=4+1=5$

Ответ: {-7; -8}, {5; 4}.

$\displaystyle \tt b). \ \ \left \{ {{y-x=2} \ \ \ \ \atop {y^{2}+4x=13}} \right.\\\\\\{} \ \ \ \ \ \left \{ {{y=x+2} \ \ \ \ \ \ \ \ \ \ \ \atop {(x+2)^{2}+4x=13}} \right.\\\\\\{} \ \ \ \ \ \ \ x^{2}+4x+4+4x=13\\\\{} \ \ \ \ \ \ \ x^{2}+8x-9=0 \ \ \ \ \ \ \ \ \ D=b^{2}-4ac=64+36=100\\\\{} \ \ \ \ \ \ \ x_{1,2}=\frac{-bб\sqrt{D}}{2a}=\frac{-8б\sqrt{100}}{2} \\\\{} \ \ \ \ \ \ \ x_{1}=-9 \ \ \ \ \ \ \ \ y_{1}=-9+2=-7 \\\\{} \ \ \ \ \ \ \ x_{2}=1 \ \ \ \ \ \ \ \ \ \ y_{2}=1+2=3$

Ответ: {-9; -7}, {1; 3}.

$\displaystyle \tt c). \ \ \left \{ {{x+y=1} \ \ \ \ \atop {x^{2}+y^{2}=25}} \right.\\\\\\{} \ \ \ \ \ \left \{ {{y=1-x} \ \ \ \ \ \ \ \ \ \ \ \atop {x^{2}+(1-x)^{2}=25}} \right.\\\\\\{} \ \ \ \ \ \ \ x^{2}+1-2x+x^{2}=25\\\\{} \ \ \ \ \ \ \ x^{2}-x-12=0 \ \ \ \ \ \ \ \ \ D=b^{2}-4ac=1+48=49\\\\{} \ \ \ \ \ \ \ x_{1,2}=\frac{-bб\sqrt{D}}{2a}=\frac{1б\sqrt{49}}{2} \\\\{} \ \ \ \ \ \ \ x_{1}=4 \ \ \ \ \ \ \ \ y_{1}=1-4=-3 \\\\{} \ \ \ \ \ \ \ x_{2}=-3 \ \ \ \ \ \ \ \ \ \ y_{2}=1+3=4$

Ответ: {4; -3}, {-3; 4}

$\displaystyle \tt d). \ \ \left \{ {{x+y=3} \ \ \ \ \atop {x^{2}+y^{2}=29}} \right.\\\\\\{} \ \ \ \ \ \left \{ {{y=3-x} \ \ \ \ \ \ \ \ \ \ \ \atop {x^{2}+(3-x)^{2}=29}} \right.\\\\\\{} \ \ \ \ \ \ \ x^{2}+9-6x+x^{2}=29\\\\{} \ \ \ \ \ \ \ x^{2}-3x-10=0 \ \ \ \ \ \ \ \ \ D=b^{2}-4ac=9+40=49\\\\{} \ \ \ \ \ \ \ x_{1,2}=\frac{-bб\sqrt{D}}{2a}=\frac{3б\sqrt{49}}{2} \\\\{} \ \ \ \ \ \ \ x_{1}=5 \ \ \ \ \ \ \ \ y_{1}=3-5=-2 \\\\{} \ \ \ \ \ \ \ x_{2}=-2 \ \ \ \ \ \ \ \ \ \ y_{2}=3+2=5$

Ответ: {5; -2}, {-2; 5}

$\displaystyle \tt51.\\\\a). \ \ \left \{ {{2x+y^{2}=6} \atop {x+y=3} \ \ \ } \right.\\\\\\{} \ \ \ \ \ \left \{ {{x=3-y} \ \ \ \ \ \ \ \ \ \ \ \atop {2\cdot (3-y)+y^{2}=6}} \right.\\\\\\{} \ \ \ \ \ \ \ y^{2}-2y+6=6\\\\{} \ \ \ \ \ \ \ y^{2}-2y=0 \ \ \ \ \ \ \ \ \\\\\ {} \ \ \ \ \ \ \ y(y-2)=0 \\\\{} \ \ \ \ \ \ \ y_{1}=0 \ \ \ \ \ \ \ \ x_{1}=3 \\\\{} \ \ \ \ \ \ \ y_{2}=2 \ \ \ \ \ \ \ \ \ \ x_{2}=3-2=1$

Ответ: {3; 0}, {1; 2}

$\displaystyle \tt b). \ \ \left \{ {{x-y=2} \ \ \ \atop {3x-y^{2}=6}} \right.\\\\\\{} \ \ \ \ \ \left \{ {{x=y+2} \ \ \ \ \ \ \ \ \ \ \ \ \atop {3\cdot (y+2)-y^{2}=6}} \right.\\\\\\{} \ \ \ \ \ \ \ -y^{2}+3y+6=6\\\\{} \ \ \ \ \ \ \ y^{2}-3y=0 \ \ \ \ \ \ \ \ \\\\\ {} \ \ \ \ \ \ \ y(y-3)=0 \\\\{} \ \ \ \ \ \ \ y_{1}=0 \ \ \ \ \ \ \ \ x_{1}=2 \\\\{} \ \ \ \ \ \ \ y_{2}=3 \ \ \ \ \ \ \ \ \ \ x_{2}=3+2=5$

Ответ: {2; 0}, {5; 3}.

$\displaystyle \tt c). \ \ \left \{ {{x-y=6} \ \ \ \atop {x^{2}+y^{2}=20}} \right.\\\\\\{} \ \ \ \ \ \left \{ {{x=y+6} \ \ \ \ \ \ \ \ \ \ \ \ \atop {(y+6)^{2}+y^{2}=20}} \right.\\\\\\{} \ \ \ \ \ \ \ y^{2}+12y+36+y^{2}=20\\\\{} \ \ \ \ \ \ \ y^{2}+6y+8=0 \ \ \ \ \ \ \ \ \ D=b^{2}-4ac=36-32=4\\\\{} \ \ \ \ \ \ \ y_{1,2}=\frac{-bб\sqrt{D}}{2a}=\frac{-6б\sqrt{4}}{2} \\\\{} \ \ \ \ \ \ \ y_{1}=-4 \ \ \ \ \ \ \ \ x_{1}=-4+6=2 \\\\{} \ \ \ \ \ \ \ y_{2}=-2 \ \ \ \ \ \ \ \ \ \ x_{2}=-2+6=4$

Ответ: {2; -4}, {4; -2}

$\displaystyle \tt d). \ \ \left \{ {{x-y=4} \ \ \ \atop {x^{2}+y^{2}=10}} \right.\\\\\\{} \ \ \ \ \ \left \{ {{x=y+4} \ \ \ \ \ \ \ \ \ \ \ \ \atop {(y+4)^{2}+y^{2}=10}} \right.\\\\\\{} \ \ \ \ \ \ \ y^{2}+8y+16+y^{2}=10\\\\{} \ \ \ \ \ \ \ y^{2}+4y+3=0 \ \ \ \ \ \ \ \ \ D=b^{2}-4ac=16-12=4\\\\{} \ \ \ \ \ \ \ y_{1,2}=\frac{-bб\sqrt{D}}{2a}=\frac{-4б\sqrt{4}}{2} \\\\{} \ \ \ \ \ \ \ y_{1}=-1 \ \ \ \ \ \ \ \ x_{1}=-1+4=3 \\\\{} \ \ \ \ \ \ \ y_{2}=-3 \ \ \ \ \ \ \ \ \ \ x_{2}=-3+4=1$

Ответ: {3; -1}, {1; -3}