\(\Leftrightarrow x+2+3⋮x+2\\ \Leftrightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-5;-3;-1;1\right\}\)

\(1,\\ a,A=\sqrt{7}-1-\dfrac{3\left(2+\sqrt{7}\right)}{-3}-2\sqrt{7}=\sqrt{7}-1+2+\sqrt{7}-2\sqrt{7}=1\\ b,B=\left(\sin^232^0+\cos^232^0\right)-\dfrac{2022\cdot\tan51^0}{\tan51^0}=1-2022=-2021\\ 2,\\ ĐK:x\ge1\\ PT\Leftrightarrow5\sqrt{x-1}-3=\sqrt{x-1}\\ \Leftrightarrow4\sqrt{x-1}=3\Leftrightarrow\sqrt{x-1}=\dfrac{3}{4}\\ \Leftrightarrow x-1=\dfrac{9}{16}\Leftrightarrow x=\dfrac{25}{16}\left(tm\right)\)

hihi lỗi kĩ thuật sorry bạn =))

\(n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ a,Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{FeCl_2}=n_{H_2}=0,1(mol)\\ a,m_{FeCl_2}=0,1.127=12,7(g)\\ b,V_{H_2}=0,1.22,4=2,24(l)\)

Gọi số giấy mỗi lớp 7A,7B,7C lần lượt là \(a,b,c(a,b,c\in \mathbb{N^*};kg)\)

Áp dụng tc dtsbn:

\(\dfrac{a}{5}=\dfrac{b}{6}=\dfrac{c}{10}=\dfrac{a+b+c}{5+6+10}=\dfrac{70}{21}=\dfrac{10}{3}\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{50}{3}\\b=20\\c=\dfrac{100}{3}\end{matrix}\right.\)

Vậy ...

\(a,\) Thay \(x=1;y=2\Leftrightarrow m-2+1=2\Leftrightarrow m=3\)

\(b,\Leftrightarrow\left\{{}\begin{matrix}m-2=2\\1\ne5\end{matrix}\right.\Leftrightarrow m=4\\ c,\text{PT giao Ox: }y=0\Leftrightarrow x=-\dfrac{1}{m-2}\Leftrightarrow A\left(-\dfrac{1}{m-2};0\right)\Leftrightarrow OA=\dfrac{1}{\left|m-2\right|}\\ \text{PT giao Oy: }x=0\Leftrightarrow y=1\Leftrightarrow B\left(0;1\right)\Leftrightarrow OB=1\\ S_{OAB}=1\Leftrightarrow\dfrac{1}{2}OA\cdot OB=1\Leftrightarrow OA\cdot OB=2\\ \Leftrightarrow\dfrac{1}{\left|m-2\right|}\cdot1=2\Leftrightarrow\left|m-2\right|=\dfrac{1}{2}\\ \Leftrightarrow\left[{}\begin{matrix}m-2=\dfrac{1}{2}\\2-m=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{5}{2}\\m=\dfrac{3}{2}\end{matrix}\right.\)

\(a,B=\dfrac{2+5}{2-3}=-7\\ b,A=\dfrac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{x-25}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ P=\dfrac{A}{B}=\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+5}=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\\ c,P< \dfrac{1}{9}\Leftrightarrow\dfrac{\sqrt{x}-5}{\sqrt{x}+3}-\dfrac{1}{9}< 0\Leftrightarrow\dfrac{8\sqrt{x}-48}{9\left(\sqrt{x}+3\right)}< 0\\ \Leftrightarrow8\sqrt{x}-48< 0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 6\Leftrightarrow0\le x< 36;x\ne9\)

Câu 43: Phản ứng hóa học nào sau đây là phản ứng phân hủy:

A. 2NaOH + Cl2 → NaCl + NaClO + H2O.

B. 2HCl + Zn(OH)2 → ZnCl2 + 2H2O.

C. CaO + CO2 → CaCO3.

D. 2KClO3 → 2KCl + 3O2↑.

Câu 44: Phản ứng hóa học nào sau đây là phản ứng hóa hợp:

A. H2SO4 + Mg → MgSO4 + H2↑

B. 2HCl + Zn(OH)2 → ZnCl2 + 2H2O

C. CaO + CO2 → CaCO3

D. 2KClO3 → 2KCl + 3O2↑

Câu 45: Phản ứng hóa học nào sau đây là phản ứng thế:

A. H2SO4 + Mg → MgSO4 + H2↑

B. 2HCl + Cu(OH)2 → CuCl2 + 2H2O

C. CaO + CO2 → CaCO3

D. 2KClO → 2KCl + O2↑ 

\(n_{CuCl_2}=0,25.0,1=0,025(mol)\\ CuCl_2+2NaOH\to Cu(OH)_2\downarrow+2NaCl\\ \Rightarrow n_{Cu(OH)_2}=0,025(mol);n_{NaOH}=0,05(mol)\\ a,m_{Cu(OH)_2}=0,025.98=2,45(g)\\ b,C_{M_{NaOH}}=\dfrac{0,05}{0,2}=0,025M\\ 2NaCl+H_2SO_4\to Na_2SO_4+2HCl\\ \Rightarrow n_{H_2SO_4}=0,025(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,025.98}{36,5\%}\approx 6,712(g)\)

Câu 1:

\(n_{HCl}=1.0,18=0,18(mol)\\ 4M+3O_2\xrightarrow{t^o}2M_2O_3\\ M_2O_3+6HCl\to 2MCl_3+3H_2O\\ \Rightarrow n_{M_2O_3}=0,03(mol)\\ \Rightarrow n_M=0,06(mol)\\ \Rightarrow M_M=\dfrac{1,62}{0,06}=27(g/mol)(Al)\\ \Rightarrow B\)

Câu 2:

\(n_{Al}=\dfrac{5,4}{27}=0,2(mol);n_{H_2SO_4}=0,3.0,1=0,03(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ LTL:\dfrac{0,2}{1}>\dfrac{0,03}{3}\Rightarrow Al\text{ dư}\\ \Rightarrow n_{H_2}=0,03(mol)\Rightarrow V_{H_2}=0,03.22,4=0,672(l)\\ n_{Al_2(SO_4)_3}=0,01(mol)\Rightarrow C_{M_{Al_2(SO_4)_3}}=\dfrac{0,01}{0,1}=0,1M\)

Chọn D