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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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=——
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=——
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=——
b-c=-1/5-1/5a=-1/5(1+a) b+c=5-5a=5(1-a)
a^2-b^2+c^2
=a^2-(b^2-c^2)
=a^2-(b-c)(b+c)
=a^2-[-1/5(1+a)]×5(1-a)
=a^2+(1+a)(1-a)
=a^2+1-a^2
=1
a消去了，得1
a²-b²+c²=a²+c²-b²=a²+(c+b)(c-b)=a²+(5-5a)(1/5-1/5a)=a²+(1-a²)=1
a²-b²+c²=a²-(12/5-13/5a)²+(3/5-12/5a)²= a^2 + (3/5-12/5a+12/5-13/5a)(3/5-12/5a-12/5+13/5a)
= a^2 + (3-5/a)(1/5a-9/5)
=a^2+ 3/5a- 1/a^2- 27/5 + 45/5a
=a^2 -1/a^2 +48/5a -27/5
....
我还以为是 12/(5a) 这样
b-c= - 1/5(1+a) ,b+c=5(1-a)
a²-b²+c²
=a²-(b²-c²)
=a²-(b+c)(b-c)
=a²-(1+a)(a-1)
=a²-a²+1
=1