\(a,=\dfrac{3y+2x}{3x^2y^2}\\ b,=\dfrac{x^2+5}{x\left(x-2\right)}\\ c,=\dfrac{6-4x+x+3}{2\left(x-3\right)\left(x+3\right)}=\dfrac{-3x+9}{2\left(x-3\right)\left(x+3\right)}\\ =\dfrac{-3\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}=\dfrac{-3}{2\left(x+3\right)}\\ d,=\dfrac{x-3+x+3}{\left(x+3\right)^2\left(x-3\right)}=\dfrac{2x}{\left(x+3\right)^2\left(x-3\right)}\)

\(\Rightarrow3x-16=\dfrac{22\cdot7^{327}}{7^{326}}=22\cdot7=154\\ \Rightarrow3x=170\\ \Rightarrow x=\dfrac{170}{3}\)

\(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)

\(d,\Rightarrow2^x\left(1+2+2^3\right)=88\\ \Rightarrow2^x\cdot11=88\Rightarrow2^x=8=2^3\Rightarrow x=3\\ e,\Rightarrow2^x=32=2^5\Rightarrow x=5\)

Bài 1:

\(=-\dfrac{1}{5}+\dfrac{4}{5}-\dfrac{13}{12}+\left(-\dfrac{1}{6}\right)\left(-\dfrac{1}{2}\right)=\dfrac{3}{5}-\dfrac{13}{12}+\dfrac{1}{12}=\dfrac{3}{5}-1=-\dfrac{2}{5}\)

Bài 2:

\(a,\Rightarrow\left[{}\begin{matrix}-4x=2\\x-\sqrt{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\sqrt{5}\end{matrix}\right.\\ b,\Rightarrow\left|2x-\dfrac{4}{5}\right|=\dfrac{4}{25}\Rightarrow\left[{}\begin{matrix}2x-\dfrac{4}{5}=\dfrac{4}{25}\\\dfrac{4}{5}-2x=\dfrac{4}{25}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{24}{25}\\2x=\dfrac{16}{25}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{12}{15}\\x=\dfrac{8}{25}\end{matrix}\right.\\ c,\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{9}\\x=-\dfrac{4}{9}\end{matrix}\right.\\ d,\Rightarrow x^3=\left(-\dfrac{2}{3}\right)^3\Rightarrow x=-\dfrac{2}{3}\\ e,\Rightarrow\left(\dfrac{4}{5}\right)^x=\left(\dfrac{4}{5}\right)^3\Rightarrow x=3\)

\(a,A=\dfrac{4-2}{4+1}=\dfrac{2}{5}\\ b,B=\dfrac{x^2-2x-x^2-2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4x-4}{\left(x-2\right)\left(x+2\right)}\\ c,P=AB=\dfrac{x-2}{x+1}\cdot\dfrac{-4\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4}{x+2}\in Z\\ \Leftrightarrow x+2\inƯ\left(-4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Leftrightarrow x\in\left\{-6;-4;-3;0\right\}\left(x\ne-1;2\right)\)

Bảo toàn KL: \(m_{MgO}=m_{Mg}+m_{O_2}=9,6+6,4=16(g)\)

Chọn C

\(M_{Fe_2O_3}=56.2+16.3=160(g/mol)\\ \%m_{Fe}=\dfrac{56.2}{160}.100\%=70\%\\ \%m_O=100\%-70\%=30\%\)

\(a,PTHH:2Al+3Cl_2\xrightarrow{t^o}2AlCl_3\\ n_{AlCl_3}=\dfrac{40,05}{133,5}=0,3(mol)\\ b,\text{Tỉ lệ: }2:3:2\\ c,n_{Al}=n_{AlCl_3}=0,3(mol)\\ \Rightarrow m_{Al}=0,3.27=8,1(g)\)

\(a,PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ b,\text{Tỉ lệ: }3:2:1\\ c,n_{Fe_3O_4}=\dfrac{1}{3}=\dfrac{1}{30}(mol)\\ \Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232\approx 7,73(g)\)

\(d,n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{1}{15}(mol)\\ \Rightarrow V_{O_2}=\dfrac{1}{15}.22,4\approx 1,49(l)\\ e,PTHH:2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ \Rightarrow n_{KMnO_4}=2n_{O_2}=\dfrac{2}{15}(mol)\\ \Rightarrow m_{KMnO_4}=\dfrac{2}{15}.158\approx 21,07(g)\)