题目内容
15.若实数a、b满足|a+2|+$\sqrt{b-4}$=0,求$\frac{{a}^{2}}{b}$的值.分析 由非负数的性质得到a+2=0,b-4=0,解得a=-2,b=4,代入求得$\frac{{a}^{2}}{b}$=1.
解答 解:∵实数a、b满足|a+2|+$\sqrt{b-4}$=0,
∴a+2=0,b-4=0,
∴a=-2,b=4,
∴$\frac{{a}^{2}}{b}$=1.
点评 本题考查了非负数的性质,算术平方根,绝对值,熟记非负数的性质是解题的关键.
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