\(a)1,5:2,16=\dfrac{15}{10}:\dfrac{216}{100}=\dfrac{15}{10}.\dfrac{100}{216}=\dfrac{5.3.10.2.5}{10.2.3.36}=\dfrac{25}{36}\\ b)\dfrac{3}{8}:0,54=\dfrac{3}{8}:\dfrac{54}{100}=\dfrac{3}{8}.\dfrac{100}{54}=\dfrac{3.4.25}{2.4.18.3}=\dfrac{25}{36}\\ c)2\dfrac{1}{3}:7=\dfrac{7}{3}.\dfrac{1}{7}=\dfrac{1}{3}\\ d)2\dfrac{2}{3}:1\dfrac{7}{9}=\dfrac{8}{3}:\dfrac{16}{9}=\dfrac{8.9}{16.3}=\dfrac{8.3.3}{8.2.3}=\dfrac{3}{2}\\ e)-0,375:3,63=-\dfrac{3}{8}:\dfrac{363}{100}=-\dfrac{3.100}{8.363}=-\dfrac{3.4.25}{4.2.3.121}=\dfrac{25}{242}\\ f)14\dfrac{2}{3}:80\dfrac{2}{3}=\dfrac{44}{3}:\dfrac{242}{3}=\dfrac{44.3}{242.3}=\dfrac{2.2.11}{2.11.11}=\dfrac{2}{11}\)

\(n_{H^+}=n_{NO_3^-}=n_{HNO_3}=0,2.4=0,8\left(mol\right)\)

PT ion rút gọn:

\(3FeS_2+12H^++15NO_3^-\rightarrow3Fe^{3+}+6SO_4^{2+}+15NO+6H_2O\)

0,1----->0,4------->0,5-------->0,1

\(\Rightarrow\left\{{}\begin{matrix}n_{H^+\left(d\text{ư}\right)}=0,8-0,4=0,4\left(mol\right)\\n_{NO_3^-\left(d\text{ư}\right)}=0,8-0,5=0,3\left(mol\right)\end{matrix}\right.\)

\(3Cu+8H^++2NO_3^-\rightarrow3Cu^{2+}+2NO+4H_2O\)

0,15<-0,4---->0,1

\(Cu+2Fe^{3+}\rightarrow Cu^{2+}+2Fe^{2+}\)

0,05<-0,1

\(\Rightarrow m=m_{Cu}=\left(0,15+0,05\right).64=12,8\left(g\right)\)

@Anguyen1127

\(\sum H=50\%.80\%=40\%\)

\(n_{CaCO_3\left(TT\right)}=\dfrac{120}{100}=1,2\left(mol\right)\\ \Rightarrow n_{CaCO_3\left(LT\right)}=\dfrac{1,2}{40\%}=3\left(mol\right)\)

PTHH:
\(\left(C_6H_{10}O_5\right)_n+nH_2O\xrightarrow[]{t^o,\text{axit}}nC_6H_{12}O_5\\ C_6H_{12}O_6\xrightarrow[]{\text{lên men}}2C_2H_5OH+2CO_2\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)

Theo PTHH:

\(n_{\left(C_6H_{10}O_5\right)_n\left(\text{cần dùng}\right)}=\dfrac{1}{n}.n_{C_6H_{12}O_6}=\dfrac{1}{2n}.n_{CO_2}=\dfrac{1}{2n}.n_{CaCO_3\left(LT\right)}=\dfrac{1,5}{n}\left(mol\right)\\ \Rightarrow m_{\left(C_6H_{10}O_5\right)_n\left(\text{cần dùng}\right)}=\dfrac{1,5}{n}.162n=243\left(g\right)\)

a)

\(2FeCO_3+4H_2SO_{4\left(\text{đ}\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+SO_2+2CO_2+4H_2O\)

\(2Fe_3O_4+10H_2SO_{4\left(\text{đ}\right)}\xrightarrow[]{t^o}3Fe_2\left(SO_4\right)_3+SO_2+10H_2O\)

b)

\(M_X=14,75.4=59\) (g/mol)

Áp dụng SĐĐC:
\(\dfrac{n_{SO_2}}{n_{CO_2}}=\dfrac{59-44}{64-59}=\dfrac{3}{1};n_X=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{SO_2}=\dfrac{3}{1+3}.0,08=0,06\left(mol\right)\\n_{CO_2}=0,08-0,06=0,02\left(mol\right)\end{matrix}\right.\)

BTNT C: \(n_{FeCO_3}=n_{CO_2}=0,02\left(mol\right)\)

BTe: \(n_{Fe_3O_4}=2n_{SO_2}-n_{FeCO_3}=0,14\left(mol\right)\)

\(\Rightarrow m=0,02.116+0,14.232=34,8\left(g\right)\)

\(M=\dfrac{x+23}{x-22}=\dfrac{x-22}{x-22}+\dfrac{45}{x-22}=1+\dfrac{45}{x-22}\)

Để M đạt GTNN thì \(1+\dfrac{45}{x-22}\) đạt GTNN

\(\Rightarrow\dfrac{45}{x-22}\) đạt GTNN \(\left(x\ne22\right)\)

\(\Rightarrow\dfrac{x-22}{45}\) đạt GTLN

\(\Rightarrow x-22\) đạt GTLN

\(\Rightarrow x-22< 0\)

Mà \(x\in Z\Rightarrow x-22=-1\Rightarrow x=-1+22=21\)

Vậy M đạt GTNN khi x = 21

\(xy+2x-y=5\\ \Leftrightarrow x.\left(y+2\right)-y-2=5-2\\ \Leftrightarrow x.\left(y+2\right)-\left(y+2\right)=3\\ \Leftrightarrow\left(x-1\right)\left(y+2\right)=3\)

Mà \(x,y\in Z\)

\(\Rightarrow\left(x-1\right)\in\text{Ư}\left(3\right)=\left\{-3;-1;1;3\right\}\)

Lập bảng

Ta có: \(\left\{{}\begin{matrix}xy=z\\yz=4x\\xz=9y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=z\\y.xy=4x\\x.xy=9y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=z\\xy^2=4x\\x^2y=9y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=z\\xy^2-4x=0\\x^2y-9y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=z\\x\left(y^2-4\right)=0\\y\left(x^2-9\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=z\\\left[{}\begin{matrix}x=0\\y=\pm2\end{matrix}\right.\\\left[{}\begin{matrix}y=0\\x=\pm3\end{matrix}\right.\end{matrix}\right.\)\(\left(\text{*}\right)\)

- Nếu \(x=0\Rightarrow\left\{{}\begin{matrix}9y=xz=0z=0\\z=yz=0y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\z=0\end{matrix}\right.\left(1\right)\)

- Nễu \(y=0\Rightarrow\left\{{}\begin{matrix}4x=yz=0z=0\\z=xy=0x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\z=0\end{matrix}\right.\left(2\right)\)

Từ \(\left(1\right),\left(2\right),\left(\text{*}\right)\Rightarrow\left[{}\begin{matrix}x=y=z=0\\\left\{{}\begin{matrix}x=\pm3\\y=\pm2\\z=xy\end{matrix}\right.\end{matrix}\right.\)

- Nếu \(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\Rightarrow z=3.2=6\)

- Nếu \(\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\Rightarrow z=\left(-3\right).2=-6\)

- Nếu \(\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\Rightarrow z=3.\left(-2\right)=-6\)

- Nếu \(\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\Rightarrow z=\left(-3\right).\left(-2\right)=6\)

Vậy \(\left(x;y;z\right)\in\left\{\left(0;0;0\right);\left(3;2;6\right);\left(-3;2;-6\right);\left(3;-2;-6\right)\left(-3;-2;6\right)\right\}\)

\(\left(x+y\right):\left(x-y\right):\left(x.y\right)=5:1:12\)

\(\Rightarrow\dfrac{x+y}{5}=\dfrac{x-y}{1}=\dfrac{xy}{12}\)

Đặt \(\dfrac{x+y}{5}=\dfrac{x-y}{1}=\dfrac{xy}{12}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=5k\left(1\right)\\x-y=k\left(2\right)\\xy=12k\end{matrix}\right.\)

Từ \(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)+\left(x-y\right)=5k+k\\\left(x+y\right)-\left(x-y\right)=5k-k\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=6k\\2y=4k\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=2k\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{xy}{y}=\dfrac{12k}{2k}=6\\y=\dfrac{xy}{x}=\dfrac{12k}{3k}=4\end{matrix}\right.\)

\(\left|x-1\right|+y^2=1\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left|x-1\right|=1\\y^2=0\end{matrix}\right.\\\left\{{}\begin{matrix}\left|x-1\right|=0\\y^2=1\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=0\\y=\pm1\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=\pm1\end{matrix}\right.\end{matrix}\right.\)

Mà \(\left|y\right|< \left|x\right|\Rightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

\(\Rightarrow\left(x+y-1\right)^{2022}=\left(2+0-1\right)^{2022}=1^{2022}=1\)

\(\text{Đ}\text{ặ}t:\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\Rightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)

Ta có: 

\(\left(x+y+z\right)^2=\left(ak+bk+ck\right)^2=\text{[}k\left(a+b+c\right)\text{]}^2=k^2\left(1\right)\)

\(x^2+y^2+z^2=\left(ak\right)^2+\left(bk\right)^2+\left(ck\right)^2=a^2k^2+b^2k^2+c^2k^2=\left(a^2+b^2+c^2\right)k^2=k^2\left(2\right)\)

Từ (1) và (2) ta có đpcm