2lg5/lg2*2lg2/lg3*2lg3/lg5=多少,

(lg3(lg2+lg5)/lg2lg5) *[ lg5(lg3+lg9)/lg3lg9]*lg2=(lg3+2lg3)/2lg3,分解因式,怎么分解的{[lg3(lg2+lg5)/lg2lg5)] * [ lg5(lg3+lg9)/lg3lg9]}*lg2 = (lg3+2lg3)/2lg3,分解因式，怎么分解的 （2lg5·2lg2·2lg3）/（lg5·lg2·lg3）= 2lg5/lg2*2lg2/lg3*2lg3/lg5=多少, 2lg2（lg2+lg5)+2lg5得数 (lg2)^2+lg2*lg5+(lg5)^2= lg2*lg50+lg5*lg20-2lg5*lg2 (lg5)平方+2*lg5*lg2+(lg2)平方 (2lg5/lg2)*(2lg2/lg3)*(2lg3/lg5)求解,谢谢!我约分完后是lg5*lg2*lg3,可是别人说等于8,请教是哪里算错了?谢谢! lg2/lg3+lg5/lg3 (lg3/2lg2+lg3/3lg2)*(lg2/lg3+lg2/2lg3) (lg2/lg3+lg2/2lg3)*(lg3/2lg2+lg3/3lg2)=? lg5/2*lg2/3*lg3/5 等于几, lg6-lg5为什么会等于lg3+2lg2-1 2^1/lg2+3^1/lg3+5^1/lg5 lg2(1+lg5)+(lg5)^2 计算：(lg5)^2 +lg2 *lg5 (lg5)^2+lg5.lg2= (lg2)^2+lg5*lg20