Question

先化简,在求值：[a^2-b^2]/[a^2-ab]÷（a＋[2ab＋b^2／a]）*(1/a+1/b),其中a=根号2+根号3,b=根号2-根号3

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Answer 1

(a²－b²)/(a²－ab)÷[a＋(2ab＋b²)/a]×(1/a＋1/b)
＝[(a＋b)(a－b)]/a(a－b)÷(a²＋2ab＋b²)/a×(a＋b)/ab
＝(a＋b)/a×a/(a＋b)²×(a＋b)/ab
＝1/ab
＝1/[(√2＋√3)(√2－√3)]
＝1/[(√2)²－(√3)²]
＝1/(2－3)
＝-1

Answer 2

[a^2-b^2]/[a^2-ab]÷（a＋[2ab＋b^2／a]）*(1/a+1/b)

=(a+b)(a-b)/a(a-b)÷(a^2+2ab+b^2)/a *(a+b)/ab
=(a+b)/a÷(a+b)^2/a *(a+b)/ab
=(a+b)/a*a/(a+b)^2*(a+b)/ab
=1/ab
=1/(√2+√3)(√2-√3)
=1/(2-3)
=-1

Answer 3

~!@#$%^&*()////////****-----+++++`

Answer 4