let a,b and c be rational numbers and b=0.24-0.26a,c=0.26-0.24a,then a^2-b^2+c^2=
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let a,b and c be rational numbers and b=0.24-0.26a,c=0.26-0.24a,then a^2-b^2+c^2=
let a,b and c be rational numbers and b=0.24-0.26a,c=0.26-0.24a,then a^2-b^2+c^2=
let a,b and c be rational numbers and b=0.24-0.26a,c=0.26-0.24a,then a^2-b^2+c^2=
原式=a²-(b²-c²)=a²-(b-c)(b+c)由于b是5分之12减去(5分之13)a,c是5分之13减去(5分之12)a,则上式=a²-(负5分之一减负五分之一a)(5-5a)=a²-【负五分之一(1+a)5(1-a)】=a²-【-(1+a)(1-a)】=a²-(a+1)(a-1)=a²-(a²-1)=a²-a²+1=1
别忘了给我加分,下面的那个人的是错的.
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证明1.Let A,B,and C be sets.Prove thatA∪包含 (A∪B ∪C).2.Let A,B,and C be sets.Prove that(A-C)∩(C -B) = 空集
let a,b and c be rational numbers and b=0.24-0.26a,c=0.26-0.24a,then a^2-b^2+c^2=
let a,b and c be rational numbers and b=0.24-0.26a,c=0.26-0.24a,then a^2-b^2+c^2=
cat( )rat A:chase B:chasing C:chased
Let a,b and c be ration numbers and b=12/5-13/5a,c=13/5-12/5a.then a2-b2+c2=?
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一.三个互不相等的有理数,即可分别表示为1,a+b,a的形式,又可分别表示为0,a/b,b的形式,则a^2000+b^2001=?二.Let a be integral part of 根号2 and b be its decimal part.Let c be the integral of π(派) and d be the decim
let a,b and c be rational numbers and b=5分之12-5分之13a,c=5分之13-5分之12a,then a的平方-b的平方+c的平方
又是一道英语的数学题.let a,b and c be rational numbers and b=12/5-13/5a,c=13/5-12/5a,then a²-b²+c²=——
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