解个六元六次方程组..2(x-u)+2λx=02(y-v)+2λ=0-2(x-u)+2ua-5a=0-2(y-v)-a=0x^2+2y-1=0u^2-5u-v+7=0谢谢..

Question

解个六元六次方程组..
2(x-u)+2λx=0
2(y-v)+2λ=0
-2(x-u)+2ua-5a=0
-2(y-v)-a=0
x^2+2y-1=0
u^2-5u-v+7=0
谢谢..

Answer 1

将这6个式子分别记为1）,2）,.6)式：
2)+4):2λ-a=0,即λ=a/2
1)+3):2λx+2ua-5a=0,代入λ,得：a(x+2u-5)=0,所以a=0 或x=5-2u
当a=0时,λ=0,由1）得：x=u,由2）y=v,代入5）u^2+2v-1=0,得：v=(1-u^2)/2,代入6）得：
u^2-5u-(1-u^2)/2+7=0,即：3u^2-10u+13=0,此方程无实根,舍去
当x=5-2u时,由5）y=(1-x^2)/2=(1-25-4u^2+20u)/2=-2u^2+10u-12
由1）5-2u-u+λ(5-2u)=0,得：λ=（3u-5)/(5-2u)
代入2）：v=y+λ=-2u^2+10u-12+(3u-5)/(5-2u)
将v代入6):u^2-5u+2u^2-10u+12+(3u-5)/(2u-5)+7=0
化为：3u^2-15u+(3u-5)/(2u-5)+19=0
6u^3-30u^2-15u^2+75u+3u-5+38u-95=0
6u^3-45u^2+116u-100=0
6u^3-12u^2-33u^2+66u+50u-100=0
(u-2)(6u^2-33u+50)=0
u=2
因此x=5-2u=1
y=-2u^2+10u-12=0
λ=（3u-5)/(5-2u)=1
v=y+λ=1
a=2λ=2