【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)_(1)}(Delta {y)_(2)}-|||-式中 Delta (y)_(1)=(y)_(1)-(y)_(e1) --底底吸收推动力;-|||-△y2=y2-y62- 塔顶吸收推动力;-|||-A=1/(m)6( 吸收因子。

8-10 对低含量气体吸收或解吸,由 dfrac (1)({k)_(y)}=dfrac (1)({k)_(y)}+dfrac (m)({k)_(x)} 出发,试

直线dfrac (x-1)(1)=dfrac (y)(2)=dfrac (z+3)(-2)-|||-__ __-|||-__与平面dfrac (x-1)(1)=

解方程。 dfrac (1)(2)x－5＝ dfrac (1)(2) 5y－2y＝18 dfrac (1)(2)＋x＝ dfrac (1)

直线dfrac (x-1)(2)=dfrac (y)(-1)=dfrac (z-2)(3)-|||-__与直线dfrac (x-1)(2)=dfrac (y)(

3、设 (x,y)=arctan dfrac (x)(y), 则 (1,1)=-|||-(A)1; (B)0; (C) dfrac {1)(2),dfrac

((y+dfrac {1)(2))}^2=dfrac (3)(4) D. ((y-dfrac {1)(2))}^2=dfrac (3)(4)

求直线dfrac (x-1)(1)=dfrac (y)(1)=dfrac (z-1)(-1)-|||-__ __在平面dfrac (x-1)(1)=dfrac

设函数 (x,y)=1-dfrac (cos sqrt {{x)^2+(y)^2}}(tan ({x)^2+(y)^2)} ,则当定设函数 (x,y)=1-df

过直线_(1):dfrac (x-1)(1)=dfrac (y-2)(0)=dfrac (z-3)(-1)-|||-__ __且平行于直线_(1):dfrac

(B) dfrac (x)(y)((y+1))^2-|||-(C) ^2((x+dfrac {1)(x))}^2. (D) dfrac (y)(x)((y+1)