两题初二数学1.化简求值1/(x+1)-[(x+3)/(x^2-1)] / [(x^2+4x+3)/(x^2-2x+1)]其中x=-3

1.解：1/(x+1)-[(x+3)/(x^2-1)] / [(x^2+4x+3)/(x^2-2x+1)]
=1/(x+1)-{(x+3)/[(x-1)(x+1)]} / {[(x+1)(x+3)]/[(x-1)^2]}
=1/(x+1)-{(x+3)[(x-1)^2]}/[(x-1)(x+1)^2(x+3)]
=1/(x+1)-(x-1)/[(x+1)^2]
=(x+1)/[(x+1)^2]-(x-1)/[(x+1)^2]
=2/[(x+1)^2]
=2/[(-3+1)^2]
=1/2
2.解：[49(a^4) -4(b^4)] / [7(a^3)b + 2a(b^3)]
={[7(a^2) -2(b^2)][7(a^2) +2(b^2)]} / {[7(a^2) + 2(b^2)]ab}
=[7(a^2) -2(b^2)]/ (ab)
=7(a/b) -2(b/a)
=7*(2/7)-2*(7/2)
=2-7
=-5

原始化简后得1/（x+1）-1/(x+1)^2
代入后得3/4

四分之三