问道初二数学题100分

来源：百度知道 编辑：UC知道 时间：2024/06/24 08:00:14
用"○""●"规定一种新运算:对于任意非零实数,都有a○b=a+1/b,
a●b=b+1/a.例如2○3=2+1/3=7/3,2●3=3+7/2,(2●3)○2=(3+1/2)○2=7/2+1/2=4. 现已知(2●x)○x=7/2,试求下列代数式的值.
(1)x^2+1/x^2 (2)x-1/x

第一问等于7，第二问等于+-根号下5，+Q374448758详细告诉你

2●x=x+1/2
(2●x)○x=(x+1/2)○x=x+1/2+1/x=7/2

so x+1/x=3

so (x+1/x)^2=x^2+1/x^2+2=3^2=9
so x^2+1/x^2=9-2=7

so (x-1/x)^2=x^2+1/x^2-2=7-2=5
so x-1/x=根号5 or -根号5

由(2●x)○x=7/2,
(x+1/2)○x=7/2,
x+1/2+1/x=7/2,
得，x+1/x=3,

所以：(1)(x+1/x)^2=9,
x^2+2+1/x^2=9,
所以x^2+1/x^2=7,

(2)x^2+1/x^2-2=5,
(x-1/x)^2=5,
所以x-1/x=±√5

(2●x)○x=7/2
(x+1/2)○x=7/2
x+1/2+1/x=7/2
x+1/x=3
(1)x^2+1/x^2=(x+1/x)^2-2=7
(2)(x-1/x)^2=(x+1/x)^2-4=5
x-1/x=+-√5

2●x=x+1/2, (2●x)○x= (x+1/2)○x= x+1/2 +1/x =7/2

x+1/x=7/2-1/2=3

(1)x^2+1/x^2 =x^2+ 2*x*1/x+ 1/x^2 - 2*x*1/x = (x+1/x)^2-2=3^2-2=7

(2) (x-1/x)^2= x^2+1/x^2 - 2*x*1/x=7-2=5

x-1/x = √5 or x-1/x = -√5

(2●x)○x=x+1/2+1/x

(1) x^2+1/x^2=(x+1/x)^2-2=(7/2-1/2)^2-2=7

(2) (x-1/x）^2= (x+1/x)^2-4
x-1/x=√5