(1/a-b - 1/a+b)÷b/a²-2ab+b² a=1+√2 b+1-√2 先化简,再求值a-b分之1减去a+b分之1除于a²-2ab+b²分之b.
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![(1/a-b - 1/a+b)÷b/a²-2ab+b² a=1+√2 b+1-√2 先化简,再求值a-b分之1减去a+b分之1除于a²-2ab+b²分之b.](/uploads/image/z/1106819-35-9.jpg?t=%281%2Fa-b+-+1%2Fa%2Bb%29%C3%B7b%2Fa%26%23178%3B-2ab%2Bb%26%23178%3B+a%3D1%2B%E2%88%9A2+b%2B1-%E2%88%9A2+%E5%85%88%E5%8C%96%E7%AE%80%2C%E5%86%8D%E6%B1%82%E5%80%BCa-b%E5%88%86%E4%B9%8B1%E5%87%8F%E5%8E%BBa%2Bb%E5%88%86%E4%B9%8B1%E9%99%A4%E4%BA%8Ea%26%23178%3B-2ab%2Bb%26%23178%3B%E5%88%86%E4%B9%8Bb.)
(1/a-b - 1/a+b)÷b/a²-2ab+b² a=1+√2 b+1-√2 先化简,再求值a-b分之1减去a+b分之1除于a²-2ab+b²分之b.
(1/a-b - 1/a+b)÷b/a²-2ab+b² a=1+√2 b+1-√2 先化简,再求值
a-b分之1减去a+b分之1除于a²-2ab+b²分之b
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(1/a-b - 1/a+b)÷b/a²-2ab+b² a=1+√2 b+1-√2 先化简,再求值a-b分之1减去a+b分之1除于a²-2ab+b²分之b.
a=1+√2
b=1-√2
(1/(a-b) - 1(/a+b)) ÷ (b/(a²-2ab+b²))
= ((a+b)-(a-b) / ((a+b)(a-b) * (a²-2ab+b²) / b
= 2b / ((a+b)(a-b) * (a-b)^2 / b
= 2(a-b) / (a+b)
= 2 {(1+√2)-(1-√2)} / ((1+√2)+(1-√2)}
= 4√2 /2
= 2√2
=[(a+b)-(a-b)]/(a+b)(a-b)*(a-b)²/b
=2(a-b)/(a+b)
把a=1+√2 b=1-√2代入得
原式=2√2
等式右侧是什么意思??
题目看不清楚,加几个括号就可以看清了
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