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已知x,y,z为实数,且1/x+1/y+1/z=2,1/x^2+1/y^2+1/z^2,则1/xy+1/yz+1/xz=
人气:152 ℃ 时间:2019-12-19 02:14:34
解答
4=(1/x+1/y+1/z)²=1/x²+1/y²+1/z²+2[1/(xy)+1/(yz)+1/(zx)],因1/x²+1/y²+1/z²的值已知,则只要代入就可以算出1/(xy)+1/(yz)+1/(zx)的值了.具体答案你的题目中的:1/x²+1/y²+1/z²的值是多少啊??等于14=(1/x+1/y+1/z)²=1/x²+1/y²+1/z²+2[1/(xy)+1/(yz)+1/(zx)],因1/x²+1/y²+1/z²=1,则:4=1+2[1/(xy)+1/(yz)+1/(zx)],得:1/(xy)+1/(yz)+1/(zx)=3/2
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