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已知|5/(x+2)-1/(y+3)|+[(y+5/(x-2)-3]^2=0,求1/3x^2·y^2的值
题目详情
已知|5/(x+2)-1/(y+3)|+[(y+5/(x-2)-3]^2=0,求1/3x^2·y^2的值
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答案和解析
|5/(x+2)-1/(y+3)|+[(y+5/(x-2)-3]^2=0,
5/(x+2)-1/(y+3)=0
y+5/(x-2)-3=0
5(y+3)=x+2
x=5y+13
y+5/(x-2)-3=0
y+5=3(x-2)
y+5=3x-6
x=y+11/3
5y+13=y+11/3
3(5y+13)=y+11
15y+39=y+11
14y=-28
y=-2
x=3
1/3x^2·y^2
=1/3*3^2*(-2)^2
=12
5/(x+2)-1/(y+3)=0
y+5/(x-2)-3=0
5(y+3)=x+2
x=5y+13
y+5/(x-2)-3=0
y+5=3(x-2)
y+5=3x-6
x=y+11/3
5y+13=y+11/3
3(5y+13)=y+11
15y+39=y+11
14y=-28
y=-2
x=3
1/3x^2·y^2
=1/3*3^2*(-2)^2
=12
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