Во втором уравнение доведем до полного квадрата:
![x^{4}+y^{4}-2x^{2}y^{2}=17-2x^{2}y^{2} \\ (x^{2}-y^{2})^{2}=17-2x^{2}y^{2} \\ 17-2x^{2}y^{2}=3^{2}=9](https://tex.z-dn.net/?f=x%5E%7B4%7D%2By%5E%7B4%7D-2x%5E%7B2%7Dy%5E%7B2%7D%3D17-2x%5E%7B2%7Dy%5E%7B2%7D+%5C%5C+%28x%5E%7B2%7D-y%5E%7B2%7D%29%5E%7B2%7D%3D17-2x%5E%7B2%7Dy%5E%7B2%7D+%5C%5C+17-2x%5E%7B2%7Dy%5E%7B2%7D%3D3%5E%7B2%7D%3D9)
![\left \{ {{x^{2}=3+y^{2}} \atop {17-2(3+y^{2})y^{2}=9} \right. \\ 17-6y^{2}-2y^{4}-9=0 \\ -2y^{4}-6y^{2}+8=0|:(-2) \\ y^{4}+3y^{2}-4=0 \\ t=y^{2} \\ t^{2}+3t-4=0](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%5E%7B2%7D%3D3%2By%5E%7B2%7D%7D+%5Catop+%7B17-2%283%2By%5E%7B2%7D%29y%5E%7B2%7D%3D9%7D+%5Cright.+%5C%5C+17-6y%5E%7B2%7D-2y%5E%7B4%7D-9%3D0+%5C%5C+-2y%5E%7B4%7D-6y%5E%7B2%7D%2B8%3D0%7C%3A%28-2%29+%5C%5C+y%5E%7B4%7D%2B3y%5E%7B2%7D-4%3D0+%5C%5C+t%3Dy%5E%7B2%7D+%5C%5C+t%5E%7B2%7D%2B3t-4%3D0+)
по теореме Виета:
![t_{1}+t_{2}=-3,t_{2}*t_{2}=-4 \\ t_{1}=-4,t_{2}=1](https://tex.z-dn.net/?f=t_%7B1%7D%2Bt_%7B2%7D%3D-3%2Ct_%7B2%7D%2At_%7B2%7D%3D-4+%5C%5C+t_%7B1%7D%3D-4%2Ct_%7B2%7D%3D1)
т.к. t=y², то не может быть отрицательным числом и корень -4 отпадает.
y²=1
![y_{1}=1,y_{2}=-1](https://tex.z-dn.net/?f=y_%7B1%7D%3D1%2Cy_%7B2%7D%3D-1)
![\left \{ {{y_{1}=1} \atop {x_{1}= +/-\sqrt{3+y^{2}}=+/- \sqrt{3+1}}=+/-2} \right. \\ \left \{ {{y_{2}=-1} \atop {x_{2}= +/-\sqrt{3+y^{2}}=+/- \sqrt{3+1}}=+/-2} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7By_%7B1%7D%3D1%7D+%5Catop+%7Bx_%7B1%7D%3D+%2B%2F-%5Csqrt%7B3%2By%5E%7B2%7D%7D%3D%2B%2F-+%5Csqrt%7B3%2B1%7D%7D%3D%2B%2F-2%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7By_%7B2%7D%3D-1%7D+%5Catop+%7Bx_%7B2%7D%3D+%2B%2F-%5Csqrt%7B3%2By%5E%7B2%7D%7D%3D%2B%2F-+%5Csqrt%7B3%2B1%7D%7D%3D%2B%2F-2%7D+%5Cright.)
и будет четыре решения:
(-2;1) (2;1) (-2;-1) (2;-1)
Я думаю, надо решать примерно так
1)
одз:
x>=-5
решаем:
![x^4*\sqrt{x+5}=0 \\x^4=0 \\x_1=0 \\\sqrt{x+5}=0 \\x+5=0 \\x_2=-5](https://tex.z-dn.net/?f=x%5E4%2A%5Csqrt%7Bx%2B5%7D%3D0+%5C%5Cx%5E4%3D0+%5C%5Cx_1%3D0+%5C%5C%5Csqrt%7Bx%2B5%7D%3D0+%5C%5Cx%2B5%3D0+%5C%5Cx_2%3D-5)
Ответ: x1=0; x2=-5
2)
одз:
![x \geq -2 \\ x \geq 6 \\ x \in [6;+\infty)](https://tex.z-dn.net/?f=x+%5Cgeq+-2+%5C%5C+x+%5Cgeq+6+%5C%5C+x+%5Cin+%5B6%3B%2B%5Cinfty%29)
решаем:
![\sqrt{x+2}=2+\sqrt{x-6} \\\sqrt{x+2}-\sqrt{x-6}=2](https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B2%7D%3D2%2B%5Csqrt%7Bx-6%7D+%5C%5C%5Csqrt%7Bx%2B2%7D-%5Csqrt%7Bx-6%7D%3D2)
возведем обе части в квадрат:
![x+2+x-6-2\sqrt{(x+2)(x-6)}=4 \\2x-4-2\sqrt{(x+2)(x-6)}=4 \\2\sqrt{(x+2)(x-6)}=2x-8 \\\sqrt{(x+2)(x-6)}=x-4](https://tex.z-dn.net/?f=x%2B2%2Bx-6-2%5Csqrt%7B%28x%2B2%29%28x-6%29%7D%3D4+%5C%5C2x-4-2%5Csqrt%7B%28x%2B2%29%28x-6%29%7D%3D4+%5C%5C2%5Csqrt%7B%28x%2B2%29%28x-6%29%7D%3D2x-8+%5C%5C%5Csqrt%7B%28x%2B2%29%28x-6%29%7D%3Dx-4)
еще раз возведем в квадрат:
![(x+2)(x-6)=(x-4)^2 \\x^2-6x+2x-12=x^2-8x+16 \\-4x-12=-8x+16 \\4x=12+16 \\4x=28 \\x=7 \in [6;+\infty)](https://tex.z-dn.net/?f=%28x%2B2%29%28x-6%29%3D%28x-4%29%5E2+%5C%5Cx%5E2-6x%2B2x-12%3Dx%5E2-8x%2B16+%5C%5C-4x-12%3D-8x%2B16+%5C%5C4x%3D12%2B16+%5C%5C4x%3D28+%5C%5Cx%3D7+%5Cin+%5B6%3B%2B%5Cinfty%29)
Ответ: x=7
1)y-x"2+2x=0
2)2X"2+Y=4X-3
3)XY=8
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