[1/(a+2)-a/(a^+3a+2)]/[2/(4-a^)]
={[(a+1)-a]/[(a+1)(a+2)]} * [(2-a)(2+a)/2]
=1/(a+1) * (2-a)/2
=(1/2)[3-(a+1)]/(a+1)
=3/[(2a+2)]-1/2
=3/[√2-1+2]-1/2
=3[√2-1]-1/2
=3√2-7/2
[(1/(a+2)-a/(a^2+3a+2)]/[2/(4-a^2)]
={(1/(a+2)-a/[(a+2)(a+1)]}/{2/[(2+a)(2-a)]}
=[1-a/(a+1)]/[2/（2-a）]
=[1/（a+1)]/[2/(2-a)]
=(2-a)/(2a+2)
=(2-sin45+sin30)/(2sin45-2sin30+2)
=(2-2^1/2*0.5+0.5)/(2*2^1/2*0.5-2*0.5+2)
=0.5*(5-2^1/2)/(2^1/2+1)
=0.5*(5-2^1/2)(2^1/2-1)
=3*2^1/2-3.5
答：先化简原式解得;-的a×9b的平方-a的平方分之6ab 再由a+b=4,a-b=2d得a=3,b=4再代入原方程又因为分 母中含有0所以原方程无解 ...详情>>
答：上当，四声。详情>>
