Nãy gõ nhầm nên xóa rồi nhé, đáp án đây:

36, C -> most

37, A -> students

38, D -> is living

39, D -> does she

40, B -> have

36, C -> most

37, A -> students

38, D -> is living in

39, D -> does she

40, B -> have

Hỏi nhẹ: Sao không tag mình vậy :)))

Bài 1:

\(\left(1\right)2NaCl\underrightarrow{dpnc}2Na+Cl_2\\ \left(2\right)Cl_2+H_2\underrightarrow{t^o}2HCl\\ \left(3\right)2HCl+Mg\rightarrow MgCl_2+H_2\\ \left(4\right)4Na+O_2\underrightarrow{t^o}2Na_2O \\ \left(5\right)Na_2O+H_2O\rightarrow2NaOH\\ \left(6\right)2Na+Cl_2\underrightarrow{t^o}2NaCl\\ \left(7\right)Na_2O+2HCl\rightarrow2NaCl+H_2O\\ \left(8\right)MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_2\)

Bài 2:

\(\left(1\right)Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ \left(2\right)Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\\ \left(3\right)2Fe_2O_3+3C\underrightarrow{t^o}4Fe+3CO_2\\ \left(4\right)Fe+2HCl\rightarrow FeCl_2+H_2\\ \left(5\right)2FeCl_2+Cl_2\underrightarrow{t^o}2FeCl_3\)

Em chỉ cần đáp án thôi ạ.

Cả đề thì nản thật, làm đến câu 12 là ngáp lên ngáp xuống, các câu sau rảnh sẽ làm tiếp

1C; 2C; 3D; 4D; 5B; 6C; 7A; 8D; 9A; 10B; 11C; 12B

làm gì thấy cái gì đâu

Với mọi x;y;z ta luôn có:

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge3\left(xy+yz+zx\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=\frac{3^2}{3}=3\)

\(B_{max}=3\) khi \(x=y=z=1\)

\(f\left(x+1\right)=\left(x+1\right)^2+p\left(x+1\right)+q\)

\(=x^2+2x+1+px+p+q\)

\(=\left(x^2+px+q\right)+2x+q+1=f\left(x\right)+2x+p+1\)

\(\Rightarrow f\left(x\right).f\left(x+1\right)=f\left(x\right)\left[f\left(x\right)+2x+p+1\right]\)

\(=f^2\left(x\right)+2x.f\left(x\right)+x^2-x^2+p.f\left(x\right)+f\left(x\right)\)

\(=\left[f\left(x\right)+x\right]^2-x^2+p.f\left(x\right)+x^2+px+q\)

\(=\left[f\left(x\right)+x\right]^2+p\left[f\left(x\right)+x\right]+q\)

\(=f\left[f\left(x\right)+x\right]\)

Thay \(x=2008\Rightarrow f\left(2008\right).f\left(2009\right)=f\left[f\left(2008\right)+2008\right]\)

Vậy tồn tại số nguyên k sao cho \(f=f\left(2008\right)+2008\) để \(f\left(k\right)=f\left(2008\right).f\left(2009\right)\)