Bài cho x,y sao lại tìm max của BT chứa a,b?

\(a,M=\dfrac{x+12+\sqrt{x}-2-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\\ b,\dfrac{1}{M}=\dfrac{\sqrt{x}-1+3}{\sqrt{x}-1}=1+\dfrac{3}{\sqrt{x}-1}\in Z\\ \Leftrightarrow\sqrt{x}-1\inƯ\left(3\right)=\left\{-1;1;3\right\}\left(\sqrt{x}-1\ge-1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;2;4\right\}\\ \Leftrightarrow x\in\left\{0;16\right\}\left(x\ne4\right)\\ c,M=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}< 1\left(\dfrac{3}{\sqrt{x}+2}>0\right)\\ d,M^2=-M\Leftrightarrow M^2+M=0\Leftrightarrow M\left(M+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}M=0\\M=-1\end{matrix}\right.\)

Với \(M=0\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow x=1\left(tm\right)\)

Với \(M=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-2\Leftrightarrow2\sqrt{x}=-1\Leftrightarrow x\in\varnothing\)

\(1,\%_{S}=\dfrac{96}{342}.100\%=\dfrac{1600}{57}\%\\ \Rightarrow m_{Al_2(SO_4)_3}=\dfrac{4,8}{\dfrac{1600}{57}\%}=17,1(g)\\ \%_{Al}=\dfrac{54}{342}.100\%=\dfrac{300}{19}\%\\ \Rightarrow m_{Al}=17,1.\dfrac{300}{19}\%=2,7(g)\\ \Rightarrow m_{S}=17,1-2,7-4,8=9,6(g)\)

\(2,\) Đặt \(n_{Al_2(SO_4)_3}=a(mol)\)

\(\Rightarrow n_{Al}=2a;n_{O}=12a(mol)\\ \Rightarrow 12a.16-27.2a=27,6\\ \Rightarrow a=0,2(mol)\\ \Rightarrow m_{O}=12.0,2.16=38,4(g)\\ m_{Al}=2.0,2.27=10,8(g)\\ m_{Al_2(SO_4)_3}=0,2.342=68,4(g)\\ \Rightarrow m_{S}=68,4-38,4-10,8=19,2(g)\)

\(A=4\sqrt{7}-3-15=4\sqrt{7}-18\\ B=\left(2\sqrt{3}+3\sqrt{3}-12\sqrt{3}\right):\sqrt{3}=-7\sqrt{3}:\sqrt{3}=-7\\ C=\sqrt{5}-\sqrt{3}-\sqrt{5}+1+\sqrt{3}=1\\ D=\left[\dfrac{\sqrt{5}\left(\sqrt{3}-2\right)}{2-\sqrt{3}}+\dfrac{\sqrt{7}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right]\left(\sqrt{7}-\sqrt{5}\right)=\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\\ E=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{\sqrt{3}-1}{1-\sqrt{3}}=\sqrt{3}+\sqrt{2}+1\\ F=\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}}-\dfrac{2\left(\sqrt{5}+2\right)}{1}+\sqrt{\left(2-\sqrt{5}\right)^2}=\sqrt{5}-2-2\sqrt{5}-4+2-\sqrt{5}=-4-2\sqrt{5}\)

Câu 5:

\(n_{Cl_2}=\dfrac{1120}{22,4}=50(mol)\\ PTHH:2KMnO_4+16HCl\to 2KCl+2MnCl_2+5Cl_2\uparrow+8H_2O\\ \Rightarrow n_{KMnO_4}=\dfrac{2}{5}n_{Cl_2}=20(mol)\\ \Rightarrow m_{KMnO_4}=20.158=3160(g)\)

Câu 6:

Đặt \(n_{Mg}=x(mol);n_{Al}=y(mol)\Rightarrow 24x+27y=10,2(1)\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5(mol)\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow x+1,5y=0,5(2)\\ (1)(2)\Rightarrow x=y=0,2(mol)\\ \Rightarrow \%_{Al}=\dfrac{0,2.27}{10,2}.100\%=52,94\%\\ \Rightarrow \%_{Mg}=100\%-52,94\%=47,06\%\\ b,\Sigma n_{HCl}=3y+2x=1(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{1}{2}=0,5(l)\)

\(n_{NaOH}=\dfrac{80}{40}=2(mol)\\ n_{Fe_2O_3}=\dfrac{0,3.10^{23}}{6.10^{23}}=0,05(mol)\)

\(\Leftrightarrow25x+24=4^5=1024\\ \Leftrightarrow25x=1000\\ \Leftrightarrow x=40\)

\(\Leftrightarrow x=\left(13+7\right):2=10\\ \Leftrightarrow y=10-7=3\)

\(a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ b,n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow n_{Fe}=0,15(mol)\\ \Rightarrow m_{Fe}=0,15.56=8,4(g)\\ c,n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,3.36,5}{36,5\%}=30(g)\)