ý ổng là e ghi sai đề đó ︵✿h̾ồn̾g̾ x̾i̾n̾h̾ g̾ái...

\(a,\Leftrightarrow\Delta=0\Leftrightarrow9-4\left(m-2\right)=0\Leftrightarrow17-4m=0\Leftrightarrow m=\dfrac{17}{4}\\ \text{Khi đó }x=-\dfrac{b}{2a}=-\dfrac{3}{2\cdot1}=-\dfrac{3}{2}\\ b,\Leftrightarrow\Delta=0\Leftrightarrow9-4\left(-2m+1\right)=0\Leftrightarrow8m+8=0\Leftrightarrow m=-1\\ \text{Khi đó }x=-\dfrac{b}{2a}=-\dfrac{3}{2\cdot1}=-\dfrac{3}{2}\\ c,\Leftrightarrow\Delta'=m^2-\left(m^2-2m+3\right)=0\Leftrightarrow2m-3=0\Leftrightarrow m=\dfrac{3}{2}\\ \text{Khi đó }x=-\dfrac{b'}{a}=-\dfrac{\dfrac{3}{2}}{1}=-\dfrac{3}{2}\\ d,\Leftrightarrow\Delta=0\Leftrightarrow\left(2m-3\right)^2-4m^2=0\Leftrightarrow-3\left(4m-3\right)=0\Leftrightarrow m=\dfrac{3}{4}\\ \text{Khi đó }x=-\dfrac{b}{2a}=\dfrac{3-2m}{2}=\dfrac{3-2\cdot\dfrac{3}{4}}{2}=\dfrac{3}{4}\)

Sửa: \(A=1+2^1+2^2+2^3+...+2^{2021}\)

\(\Rightarrow A+1=1+1+2+2^2+...+2^{2021}\\ \Rightarrow A+1=2+2+2^2+...+2^{2021}\\ \Rightarrow A+1=2^2+2^2+2^3+...+2^{2021}\\ \Rightarrow A+1=2^3+2^3+2^4+...+2^{2021}\\ ....\\ \Rightarrow A+1=2^{2021}+2^{2021}=2^{2022}\)

Mà \(2^x=A+1\Rightarrow2^x=2^{2022}\Rightarrow x=2022\)

\(=2\sqrt{5}+4\sqrt{5}-3\sqrt{5}=3\sqrt{5}\)

\(=\left(x^3-2x^4+3x-6x^2-1+2x\right):\left(1-2x\right)\\ =\left[x^3\left(1-2x\right)+3x\left(1-2x\right)-\left(1-2x\right)\right]:\left(1-2x\right)\\ =x^3+3x-1\)

\(n_{O_2}=0,3(mol)\Rightarrow n_{O}=0,3.2=0,6(mol)\)

Bảo toàn C và H: \(n_C=n_{CO_2}=0,2(mol);n_H=2n_{H_2O}=0,3.2=0,6(mol)\)

Đặt \(CTHH_A:C_xH_yO_z\)

\(\Rightarrow x:y:z=0,2:0,6:0,6=1:3:3\\ \Rightarrow CTHH_A:CH_3O_3\)

Đề sai

\(a,2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ FeCl_3+3NaOH\to Fe(OH)_3\downarrow+3NaCl\\ 2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ Fe_2(SO_4)_3+3Zn\to 3ZnSO_4+2Fe\\ Fe+2HCl\to FeCl_2+H_2\\ b,4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ Al_2O_3\xrightarrow[cryolit]{đpnc}4Al+3O_2\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3\downarrow+3Na_2SO_4\\ 2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\)

Gọi số cây 7A,7B,7C lần lượt là \(a,b,c(a,b,c\in \mathbb{N^*})\)

Áp dụng tc dtsbn:

\(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=\dfrac{a+b+c}{3+4+5}=\dfrac{240}{12}=20\\ \Leftrightarrow\left\{{}\begin{matrix}a=60\\b=80\\c=100\end{matrix}\right.\)

Vậy ...

\(2Mg+O_2\xrightarrow{t^o}2MgO\\ Mg+2HCl\to MgCl_2+H_2\\ MgCl_2+2NaOH\to Mg(OH)_2+2NaCl\)

Do đó: \(\begin{cases} X,Z:Mg\\ Y:MgO\\ T:MgCl_2\\ U:H_2\\ F:NaCl\\ V:Mg(OH)_2 \end{cases}\)

Ta có:

\(n_{Mg}=\dfrac{a}{24}\Rightarrow n_{MgO}=\dfrac{a}{48}\\ \Rightarrow m_1=m_{MgO}=\dfrac{a}{48}.40=\dfrac{5a}{6}\\ n_{MgCl_2}=\dfrac{a}{48}\Rightarrow n_{Mg(OH)_2}=\dfrac{a}{48}\\ \Rightarrow m_2=m_{Mg(OH)_2}=\dfrac{a}{48}.58=\dfrac{29a}{24}\\ \Rightarrow \dfrac{m_1}{m_2}=\dfrac{5}{6}:\dfrac{29}{24}=\dfrac{20}{29}\)

tính tắt đấy bạn

mẫu số là khối lượng dd sau p/ứ

\(m_{dd_{KCl}}=m_{K_2CO_3}+m_{dd_{HCl}}-m_{CO_2}=17,25+50-0,125.44=...\)

Ko tính H₂O nhé

Còn tử số là khối lượng chất tan của KCl

\(m_{CT_{KCl}}=0,25.74,5\)