\(sin16^0=sin\left(90^0-74^0\right)=cos74^0\)

\(sin60^0=sin\left(90^0-60^0\right)=cos30^0\)

Sắp xếp: \(cos16^0,cos30^0,cos43^0,cos52^0,cos74^0\)

\(A\cap B=\varnothing\)\(\Leftrightarrow\left[{}\begin{matrix}m\ge3\\m+1< 0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m\ge3\\m< -1\end{matrix}\right.\)

Ta có nên làm bánh mì sữa không, ngồi trong phòng mọc nấm mất... Nhưng mà hong có men, haizz, lười quá

Để \(B\subset A\Leftrightarrow3m+3< m\Leftrightarrow m< -\dfrac{3}{2}\)

Để \(B\subset A\) \(\Leftrightarrow m-2\le-1\Leftrightarrow m\le1\)

Để \(B\subset A\) \(\Leftrightarrow4\le m+1\) \(\Leftrightarrow m\ge3\)

Để \(B\subset A\)\(\Leftrightarrow\left\{{}\begin{matrix}2m-3>-1\\2m< 4\end{matrix}\right.\)

\(\Leftrightarrow1< m< 2\)

Để \(B\subset A\Leftrightarrow\)\(\left\{{}\begin{matrix}m-7\ge-4\\m\le3\end{matrix}\right.\)

\(\Leftrightarrow m=3\)

Đây iem ey:

\(\dfrac{x+8}{x+5}=\dfrac{x+5+3}{x+5}=1+\dfrac{3}{x+5}\in Z\)

Khi \(x+3\) thuộc ước của 5 hay 3 chia hết cho 5

Đó, làm như trên (o=o) 

\(n_{Fe}=0,1\left(mol\right)\)

 \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0,1\)       \(0,2\)          \(0,1\)               

\(m_{ddHCl}=\dfrac{0,2.36,5}{50\%}=36,5\left(g\right)\)

\(m_{H_2}=0,1.2=0,2\left(g\right)\)

\(m_{dd}=5,6+36,5-0,2=41,9\left(g\right)\)

\(C\%\left(FeCl_2\right)=\dfrac{0,1.127}{41,9}=30,3\left(\%\right)\)

\(\dfrac{360}{x}-\dfrac{400}{x+1}=1\)

Điều kiện: \(x\ne0;x\ne-1\)

PT \(\Leftrightarrow\dfrac{360\left(x+1\right)-400x}{x\left(x+1\right)}=1\)

\(\Rightarrow-40x+360=x\left(x+1\right)\)

\(\Leftrightarrow-40x+360=x^2+x\)

\(\Leftrightarrow x^2+41x-360=0\)

\(\Leftrightarrow x^2+2.\dfrac{41}{2}.x+\dfrac{1681}{4}=\dfrac{3121}{4}\)

\(\Leftrightarrow\left(x+\dfrac{41}{2}\right)^2=\left(\dfrac{\sqrt{3121}}{2}\right)^2\)

\(\Leftrightarrow x+\dfrac{41}{2}=\dfrac{\sqrt{3121}}{2}\) hoặc \(x+\dfrac{41}{2}=-\dfrac{\sqrt{3121}}{2}\)

\(\Leftrightarrow x=\dfrac{\sqrt{3121}}{2}-\dfrac{41}{2}\) hoặc \(x=-\dfrac{\sqrt{3121}}{2}-\dfrac{41}{2}\)

Vậy...