已知(x-x^2)+(x^2-y)=1,求代数式1/2(x^2+y^2)-xy的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 17:26:06
![已知(x-x^2)+(x^2-y)=1,求代数式1/2(x^2+y^2)-xy的值](/uploads/image/z/3757249-1-9.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%88x-x%5E2%EF%BC%89%2B%28x%5E2-y%29%3D1%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F1%2F2%28x%5E2%2By%5E2%29-xy%E7%9A%84%E5%80%BC)
x){}KЭ3zS[HVji"}
M/+40u+t+,4`1[
&LXרREhN!$A`虂ulyo:ro>`OVgodR0
已知(x-x^2)+(x^2-y)=1,求代数式1/2(x^2+y^2)-xy的值
已知(x-x^2)+(x^2-y)=1,求代数式1/2(x^2+y^2)-xy的值
已知(x-x^2)+(x^2-y)=1,求代数式1/2(x^2+y^2)-xy的值
(x-x^2)+(x^2-y)=1
x-y=1
1/2(x^2+y^2)-xy
=(1/2)(x^2+y^2-2xy)
=(1/2)(x-y)^2
=(1/2)*1^2
=1/2
0.5
∴得x-y=1
∴原式=(x-y)^2/2=0.5