Đáp án:

\(\root 3 \of 4 – \root 3 \of 3 \)

Giải thích các bước giải:

$$\eqalign{
& A = {1 \over {\root 3 \of {16} + \root 3 \of {12} + \root 3 \of 9 }} \cr
& = {1 \over {{{\left( {\root 3 \of 4 } \right)}^2} + \root 3 \of 4 .\root 3 \of 3 + {{\left( {\root 3 \of 3 } \right)}^2}}} \cr
& = {{\root 3 \of 4 – \root 3 \of 3 } \over {\left( {\root 3 \of 4 – \root 3 \of 3 } \right)\left[ {{{\left( {\root 3 \of 4 } \right)}^2} + \root 3 \of 4 .\root 3 \of 3 + {{\left( {\root 3 \of 3 } \right)}^2}} \right]}} \cr
& = {{\root 3 \of 4 – \root 3 \of 3 } \over {{{\left( {\root 3 \of 4 } \right)}^3} – {{\left( {\root 3 \of 3 } \right)}^3}}} = {{\root 3 \of 4 – \root 3 \of 3 } \over {4 – 3}} = \root 3 \of 4 – \root 3 \of 3 \cr} $$

FeO + 2HCl -> FeCl2 +H2O

x 2x x

CuO + 2HCl -> CuCl2 + H2O

y 2y y

nHCl= 100.29,2% /36,5 = 0,8

72x+80y=30,4

2x+2y= 0,8

⇒x=y=0,2

⇒%mFeO= 0,2.72 / 30,4 .100% = 47,37%

⇒%mCuO= 52,63%

mdd sau phản ứng = 100+30,4=130,4

⇒C% FeCl2 =0,2 . 127 / 130,4 .100% = 19,48%

⇒C%= 0,2.135/ 130,4 .100%=20,71%