理想气体内能增量Delta E=dfrac (m)(M)dfrac (i)(2)RDelta T,由于Delta E=dfrac (m)(M)dfrac (i)(2)RDelta T,故Delta E=dfrac (m)(M)dfrac (i)(2)RDelta T,该方程适用于Delta E=dfrac (m)(M)dfrac (i)(2)RDelta TDelta E=dfrac (m)(M)dfrac (i)(2)RDelta T等体过程; Delta E=dfrac (m)(M)dfrac (i)(2)RDelta T等压过程;Delta E=dfrac (m)(M)dfrac (i)(2)RDelta T绝热过程; Delta E=dfrac (m)(M)dfrac (i)(2)RDelta T任何过程

=dfrac (T)(2pi )leqslant [ t] =dfrac ([ O] )(2);-|||-_(2)(A)_(1)+dfrac ({M)_(y)}

(4)设 (x)=(e)^sqrt (x) 则 lim _(Delta xarrow 0)dfrac (f(1+Delta x)-f(1))(Delta x)=

【题 7-14] 对于低浓度气体的逆流吸收,试证明:-|||-_(OC)=dfrac (1)(1-dfrac {1)(A)}ln dfrac (Delta {y

在 Delta ABC 中, angle A=dfrac (pi )(6),在 Delta ABC 中, angle A=dfrac (pi )(6),

[ dfrac {{e)^2}({(z-1))^2},1] =-|||-A e-|||-B dfrac (1)(e)-|||-C dfrac (e)(2)-||

△z=3m =m(2300-3230)-(10)^3-dfrac (1)(10)-dfrac (10)(3600) .q=0-|||-=2mdfrac ({E)

(a)_(m)+(b)_(1)+(b)_(2)+... +(b)_(n)}(m+n)=-|||-dfrac (m)(m+n)cdot dfrac ({a)_(1

πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T

=dfrac ({n)^2h}(8m/{s)^2}-|||-B omega =(x)=sqrt (dfrac {2)(t)}sin (dfrac ({n)^2p

f(x)= ({e)^4-dfrac (1)(3))-|||-dfrac (1)(2)(e)^4-|||-dfrac (1)(2)((e)^2-dfrac