$412+123-12$

$=(412-12)+123$

$=400+123=523$

Mình cảm ơn bạn nhé !!!

\(7+\left(\dfrac{7}{12}-\dfrac{1}{2}+3\right)-\left(\dfrac{1}{12}+5\right)\)

\(=7+\left(\dfrac{7}{12}-\dfrac{6}{12}+3\right)-\left(\dfrac{1}{12}+5\right)\)

\(=7+3+\dfrac{1}{12}-\dfrac{1}{12}-5\)

=10-5=5

= 5

\(\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+2\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\\ =\left(\sqrt{5}+2\right)+\left(\sqrt{3}+1\right)-\left(\sqrt{5}+\sqrt{3}\right)\\ =\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}\\ =2+1=3\)

\(-\dfrac{1}{12}-\left(-\dfrac{1}{10}\right)\\ =-\dfrac{1}{12}+\dfrac{1}{10}\\ =\dfrac{-5}{60}+\dfrac{6}{60}\\ =\dfrac{-5+6}{60}\\ =\dfrac{1}{60}\)

-1/12--1/10

=-5/60--6/60

=-11/60

$2,5\times20,21\times5\times40\times0,2$

$=(2,5\times40)\times(5\times0,2)\times20,21$

$=100\times1\times20,21$

$=100\times20,21=2021$

\(2,5\cdot20,21\cdot5\cdot40\cdot0,2\\=\left(2,5\cdot40\right)\cdot20,21\cdot\left(5\cdot0,2\right)\\ =\left(2,5\cdot4\cdot10\right)\cdot20,21\cdot1\\ =100\cdot20,21\\ =2021\)

\(\sqrt{\dfrac{9}{4}}-\sqrt{2}+\sqrt{2}\\ =\dfrac{3}{2}-\left(\sqrt{2}-\sqrt{2}\right)\\ =\dfrac{3}{2}-0\\ =\dfrac{3}{2}\)

\(\left(7-\dfrac{1}{5}+\dfrac{1}{3}\right)-\left(6+\dfrac{9}{5}+\dfrac{4}{3}\right)\\ =7-\dfrac{1}{5}+\dfrac{1}{3}-6-\dfrac{9}{5}-\dfrac{4}{3}\\ =\left(7-6\right)-\left(\dfrac{1}{5}+\dfrac{9}{5}\right)+\left(\dfrac{1}{3}-\dfrac{4}{3}\right)\\ =1-2-1\\ =-2\)

\(\left(7-\dfrac{1}{5} +\dfrac{1}{3}\right)-\left(6+\dfrac{9}{5}+\dfrac{4}{3}\right)\)

\(=7-\dfrac{1}{5}+\dfrac{1}{3}-6-\dfrac{9}{5}-\dfrac{4}{3}\)

\(=\left(7-6\right)-\left(\dfrac{1}{5}+\dfrac{9}{5}\right)+\left(\dfrac{1}{3}-\dfrac{4}{3}\right)-\dfrac{1}{3}\)

\(=1-2+\left(-1\right)-\dfrac{1}{3}\)

\(=\left[1+\left(-1\right)\right]-2-\dfrac{1}{3}\)

\(=0-2-\dfrac{1}{3}\)

\(=-2-\dfrac{1}{3}\)

\(=-\dfrac{6}{3}-\dfrac{1}{3}\)

\(=-\dfrac{7}{3}\)

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.\left(2^2.5\right)}\)

\(=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\\ =\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\\ =\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\\ =\dfrac{-2}{6}\\ =-\dfrac{1}{3}\)

\(H=\dfrac{4}{1-\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\\ =\dfrac{4\left(1+\sqrt{3}\right)}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\\ =\dfrac{4\left(1+\sqrt{3}\right)}{1-3}-\sqrt{3}\\ =\dfrac{4\left(1+\sqrt{3}\right)}{-2}-\sqrt{3}\\ =-2\left(1+\sqrt{3}\right)-\sqrt{3}\\ =-2-2\sqrt{3}-\sqrt{3}\\ =-2-3\sqrt{3}\)

Cảm ơn bạn nhé

\(n_{H_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)

PTHH:

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 

x        2x                 x           x 

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

y             2y            y            y 

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

z           2z              z              z 

\(ZnCl_2+2NaOH\rightarrow2NaCl+Zn\left(OH\right)_2\downarrow\)

x                2x                2x               x

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

y                 2y                  y                2y

\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)

z                    2z                 z                   2z

\(Zn\left(OH\right)_2\rightarrow\left(t^o\right)ZnO+H_2O\)

x                             x             x 

\(4Fe\left(OH\right)_2+O_2\rightarrow\left(t^o\right)2Fe_2O_3+4H_2O\)

y                               0,5y             y 

\(Mg\left(OH\right)_2\rightarrow\left(t^o\right)MgO+H_2O\)

z                             z                  z

Ta có Hệ PT:

\(\left\{{}\begin{matrix}x+y+z=0,5\\65x+56y+24z=20,9\\81x+80y+40z=22\end{matrix}\right.\)

Bạn check lại đề có sai ở đâu k nha chứ bậm nghiệm k ra.

Ta có: 65n_Zn + 56n_Fe + 24n_Mg = 20,9 (1)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{Zn}+n_{Fe}+n_{Mg}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\left(2\right)\)

PT: \(ZnCl_2+2NaOH\rightarrow Zn\left(OH\right)_2+2NaCl\)

\(Zn\left(OH\right)_2+2NaOH\rightarrow Na_2ZnO_2+2H_2O\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_{2\downarrow}+2NaCl\)

\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_{2\downarrow}+2NaCl\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

Theo PT: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{FeCl_2}=\dfrac{1}{2}n_{Fe}\\n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}=n_{Mg}\end{matrix}\right.\)

\(\Rightarrow m_{ran}=m_{Fe_2O_3}+m_{MgO}=160.\dfrac{1}{2}n_{Fe}+40n_{Mg}=22\left(g\right)\left(3\right)\)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,1\left(mol\right)\\n_{Fe}=0,15\left(mol\right)\\n_{Mg}=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1.65=6,5\left(g\right)\\m_{Fe}=0,15.56=8,4\left(g\right)\\m_{Mg}=0,25.24=6\left(g\right)\end{matrix}\right.\)