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(x-y)²=12,(x+y)=16,求x²+y²的值
人气:166 ℃ 时间:2020-04-22 19:15:24
解答
(x-y)²=12,(1)
(x+y)=16
两边平方得:
(x+y)²=256(2)
(1)+(2)得:
2(x²+y²)=12+256
2(x²+y²)=268
x²+y²=268÷2=134(x-y)²=12,(x+y)²=16,求x²+y²的值不好意思发错了(x-y)²=12,(1)(x+y)=16 (2)(1)+(2)得:2(x²+y²)=12+162(x²+y²)=28x²+y²=28÷2=14
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