已知根号(2x-y-4)+根号(x-2y-5)=0,求根号(y^2+5x)的值
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已知根号(2x-y-4)+根号(x-2y-5)=0,求根号(y^2+5x)的值
已知根号(2x-y-4)+根号(x-2y-5)=0,求根号(y^2+5x)的值
已知根号(2x-y-4)+根号(x-2y-5)=0,求根号(y^2+5x)的值
根号运算得到的结果必然是0或正数,因此要满足题中等式,则必须有
2x-y-4=0
x-2y-5=0
解得x=1 y=-2
故sqrt(y^2+5x) = sqrt(9) = 3
uuuy
ii
2x-y-4=0
x-2y-5=0
x=1
y=-2
y^2+5x=9
由√(2x-y-4)+√(x-2y-5)=0 2x-y-4=0 x-2y-5=0
解得 x=1 y=-2
所以√(y^2+5x)=3
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