已知(x-y)²=4,(x+y)²=64;求下列代数式的值 (1)x²+y²(2)xy
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/25 17:09:59
![已知(x-y)²=4,(x+y)²=64;求下列代数式的值 (1)x²+y²(2)xy](/uploads/image/z/3168144-0-4.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%88x-y%EF%BC%89%26%23178%3B%3D4%2C%EF%BC%88x%2By%EF%BC%89%26%23178%3B%3D64%EF%BC%9B%E6%B1%82%E4%B8%8B%E5%88%97%E4%BB%A3%E6%95%B0%E5%BC%8F%E7%9A%84%E5%80%BC+%EF%BC%881%EF%BC%89x%26%23178%3B%2By%26%23178%3B%EF%BC%882%EF%BC%89xy)
已知(x-y)²=4,(x+y)²=64;求下列代数式的值 (1)x²+y²(2)xy
已知(x-y)²=4,(x+y)²=64;求下列代数式的值 (1)x²+y²(2)xy
已知(x-y)²=4,(x+y)²=64;求下列代数式的值 (1)x²+y²(2)xy
1、x²+y²=(x+y)²+(x-y)²
=4+64
=68
2、xy=[(x+y)²-(x-y)²]/4
=(64-4)/4
=15
(x-y)²=4①
(x+y)²=64②
由①-②:-4xy=-60 xy=15
(1)x²+y²=(x+y)²-2xy=64-30=14
(2)xy=15
(1)x²+y²=[(x-y)²+(x+y)²]/2=34
(2)xy=[(x+y)²+(x-y)²]/4=15
(x-y)^2+(x+y)^2=2(x^2+y^2)=68,所以x^2+y^2=34; (x+y)^2-(x-y)^2=4xy=60.所以xy=15