Question

已知a+2b=0,求(a²+2ab-b²)÷(2a²+ab+b²)的值

Answer 1

a+2b=0
∴a=-2b
把a=-2b代入(a²+2ab-b²)÷(2a²+ab+b²)得
(a²+2ab-b²)÷(2a²+ab+b²)
=(4b²-4b²-b²)÷(8b²-2b²+b²)
=-b²/7b²
=-1/7

Answer 2

a+2b=0
a=-2b
a/b=-2

(a²+2ab-b²)÷(2a²+ab+b²)分子分母同时除以b²
=(a²/b²+2ab/b²-b²/b²)÷(2a²/b²+ab/b²+b²/b²)
=(a²/b²+2a/b-1)÷(2a²/b²+a/b+1)
=[(-2)²+2*(-2)-1]÷[2*(-2)²-2+1]
=[4-4-1]÷[8-2+1]
=-1÷7
=-1/7

Answer 3

a+2b=0, a=-2b
(a²+2ab-b²)÷(2a²+ab+b²)
=[4b²-4b²-b²]÷(8b²-2b²+b²)
=(-b²)÷(7b²)
=-1/7

Answer 4

a=-2b
[(-2b)^2-4b^2-b^2]/[2(-2b)^2-2b^2+b^2]=-1/7

Answer 5

a,b 有什么条件没：
1：a=b=0 后面的式子错误，0/0
2：a,b不为0，a=-2b,带入后面式子，化简计算。结果为-1/7