Làm trong 21 ngày cần \(15\times35:21=25\left(máy\right)\)

Vậy cần thêm \(25-15=10\left(máy\right)\)

\(b,\left(d_3\right)\text{//}\left(d_1\right)\Leftrightarrow\left\{{}\begin{matrix}a=1\\b\ne4\end{matrix}\right.\Leftrightarrow\left(d_3\right):y=x+b\)

PT hoành độ giao điểm \(\left(d_2\right);\left(d_3\right)\) là \(x+b=-2x-2\)

Mà 2 đt cắt tại hoành độ \(-3\) nên \(x=-3\)

\(\Leftrightarrow b-3=4\Leftrightarrow b=7\)

Vậy \(\left(d_3\right):y=x+7\)

\(a,\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x-y}{7-4}=\dfrac{12}{3}=4\\ \Rightarrow\left\{{}\begin{matrix}x=4.7=28\\y=4.4=16\end{matrix}\right.\\ b,\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{2}=\dfrac{z}{2}\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{6}=\dfrac{x+y+z}{4+6+6}=\dfrac{50}{16}=\dfrac{25}{8}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{25}{8}.4=\dfrac{25}{2}\\y=\dfrac{25}{8}.6=\dfrac{75}{4}\\z=\dfrac{25}{8}.6=\dfrac{75}{4}\end{matrix}\right.\)

Câu c:

\(M=\dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{z+t+x}\\ \Rightarrow M>\dfrac{x}{x+y+z+t}+\dfrac{y}{x+y+z+t}+\dfrac{z}{x+y+z+t}+\dfrac{t}{x+y+z+t}=1\)

Mà \(M< \dfrac{x+t}{x+y+z+t}+\dfrac{y+z}{x+y+z+t}+\dfrac{z+x}{x+y+z+t}+\dfrac{y+t}{x+y+z+t}=\dfrac{2\left(x+y+z+t\right)}{x+y+z+t}=2\)

Do đó \(1< M< 2\Rightarrow M\notin N\)

Câu a:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2k^2+bdk^2}{d^2k^2-bdk^2}=\dfrac{bk^2\left(b+d\right)}{dk^2\left(d-b\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\\\dfrac{b^2+bd}{d^2-bd}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\end{matrix}\right.\\ \RightarrowĐpcm\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{Fe}=n_{H_2SO_4}=n_{Fe(OH)_2}=0,2(mol)\\ a,m_{Fe}=0,2.56=11,2(g)\\ b,C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1M\)

\(c,Ba(OH)_2+FeSO_4\to BaSO_4\downarrow+Fe(OH)_2\downarrow\\ n_{Ba(OH)_2}=\dfrac{250.17,1}{100.171}=0,25(mol)\\ LTL:\dfrac{0,2}{1}<\dfrac{0,25}{1}\Rightarrow Ba(OH)_2\text{ dư}\\ \Rightarrow n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{BaSO_4}=233.0,2=46,6(g)\)

Câu 1:

\(CTHH_A:T_2O_3\\ \Rightarrow M_A=\dfrac{16.3}{47\%}=102(g/mol)\\ \Rightarrow M_T=\dfrac{102-3.16}{2}=27(g/mol)\\ \Rightarrow T(nhôm,Al)\\ \Rightarrow CTHH_A:Al_2O_3\)

Câu 2:

\(\%_H=100\%-82,35\%=17,65\%\\ CTHH_A:XH_3\\ \Rightarrow M_A=\dfrac{3}{17,65\%}=17(g/mol)\\ \Rightarrow M_X=17-3=14(g/mol)\\ \Rightarrow X(natri,Na)\\ \Rightarrow CTHH_A:NH_3\)

\(a,CTTQ:Na_x^IO_y^{II}\\ \Rightarrow x.I=y.II\Rightarrow \dfrac{x}{y}=2\Rightarrow x=2;y=1\\ \Rightarrow CTHH:Na_2O\\ b,CTTQ:Cu_x^{II}O_y^{II}\\ \Rightarrow x.II=y.II\Rightarrow \dfrac{x}{y}=1\Rightarrow x=1;y=1\\ \Rightarrow CTHH:CuO\)

\(c,CTTQ:Al_x^{III}(OH)_y^{I}\\ \Rightarrow x.III=y.I\Rightarrow \dfrac{x}{y}=\dfrac{1}{3}\Rightarrow x=1;y=3\\ \Rightarrow CTHH:Al(OH)_3\\ d,CTTQ:Ca_x^{II}(PO_4)_y^{III}\\ \Rightarrow x.II=y.III\Rightarrow \dfrac{x}{y}=\dfrac{3}{2}\Rightarrow x=3;y=2\\ \Rightarrow CTHH:Ca_3(PO_4)_2\)

\(4P+5O_2\xrightarrow{t^o}2P_2O_5\\ n_P=\dfrac{6,2}{31}=0,2(mol)\\ \Rightarrow n_{P_2O_5}=0,1(mol);n_{O_2}=0,25(mol)\\ \Rightarrow m_{P_2O_5}=0,1.142=14,2(g);V_{O_2}=0,25.22,4=5,6(l)\)

\(n_{H_2}=\dfrac{3,24}{24}=0,135(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{H_2SO_4}=n_{FeSO_4}=0,135(mol)\\ \Rightarrow \begin{cases} C_{M_{H_2SO_4}}=\dfrac{0,135}{0,2}=0,675M\\ C_{M_{FeSO_4}}=\dfrac{0,135}{0,2}=0,675M \end{cases}\)