\(\frac{4+\sqrt{X}}{7}\)

Gọi x là số giờ làm khẩu trang

Gọi y là số khẩu trang làm trong 1 giờ \(\left(ĐK:x;y>0\right)\) 

Theo đề, ta có 

\(\hept{\begin{cases}xy=400\\\frac{1}{2}xy+\left(\frac{1}{2}x-1\right)\left(y+100\right)=400\end{cases}}\)   

\(\hept{\begin{cases}xy=400\\\frac{1}{2}xy+50x-y-100=200\end{cases}}\)   

\(\hept{\begin{cases}xy=400\\50x-y=100\end{cases}}\)   

\(\hept{\begin{cases}y=\frac{400}{x}\\50x-\frac{400}{x}=100\end{cases}}\)   

\(\hept{\begin{cases}y=\frac{400}{x}\\50x^2-100x-400=0\end{cases}}\)    

\(\hept{\begin{cases}y=\frac{400}{x}\\x^2-2x-8=0\end{cases}}\)    

\(\hept{\begin{cases}y=\frac{400}{x}\\x=4\left(n\right);x=-2\left(l\right)\end{cases}}\)   

\(\hept{\begin{cases}y=100\\x=4\end{cases}}\)

\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{9-2.3.2\sqrt{2}+8}+\sqrt{8+2.2\sqrt{2}+1}\\ \)

\(=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =\left|3-2\sqrt{2}\right|+\left|2\sqrt{2}-1\right|\\ =3-2\sqrt{2}+2\sqrt{2}+1\\ =4\)

Ta có \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\left(x,y,z>0\right)\).

\(\Leftrightarrow\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\).

\(P=\frac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+y^2}\right)\)\(\left(x,y,z>0\right)\).

Ta có: 

\(\sqrt{2y^2+2yz+2z^2}=\sqrt{\frac{5}{4}\left(y^2+2yz+z^2\right)+\frac{3}{4}\left(y^2-2yz+z^2\right)}\)

\(=\sqrt{\frac{5}{4}\left(y+z\right)^2+\frac{3}{4}\left(y-z\right)^2}\).

Ta có:

\(\frac{3}{4}\left(y-z\right)^2\ge0\forall y;z>0\).

\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2\ge\frac{5}{4}\left(y+z\right)^2\forall y;z>0\).

\(\Rightarrow\sqrt{\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y,z>0\).

\(\Leftrightarrow\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y;z>0\).

\(\Leftrightarrow x\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}x\left(y+z\right)\forall x;y;z>0\left(1\right)\).

Chứng minh tương tự, ta được:

\(y\sqrt{2x^2+xz+2z^2}\ge\frac{\sqrt{5}}{2}y\left(x+z\right)\forall x;y;z>0\left(2\right)\).

Chứng minh tương tự, ta được:

\(z\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}z\left(x+y\right)\forall x;y;z>0\left(3\right)\).

Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:

\(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+2y^2}\)\(\ge\)\(\frac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]=\sqrt{5}\left(xy+yz+zx\right)\).

\(\Leftrightarrow\frac{1}{xyz}\left(x\sqrt{2y^2+yz+z^2}+y\sqrt{2z^2+zx+2x^2}+z\sqrt{2x^2+xy+2y^2}\right)\)\(\ge\)\(\frac{\sqrt{5}\left(xy+yz+zx\right)}{xyz}=\sqrt{5}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\).

\(\Leftrightarrow P\ge\frac{\sqrt{5}}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\)

\(\left(4\right)\).

Vì \(x,y,z>0\)nên áp dụng bất đẳng thức Bu-nhi-a-cốp-xki, ta được:
\(\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\)\(\left(1.\frac{1}{\sqrt{x}}+1.\frac{1}{\sqrt{y}}+1.\frac{1}{\sqrt{z}}\right)^2\).

\(\Leftrightarrow\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2=1^2=1\)

(vì\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\)).

\(\Leftrightarrow\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\frac{\sqrt{5}}{3}\)\(\left(5\right)\).

Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:

\(P\ge\frac{\sqrt{5}}{3}\).

Dấu bằng xảy ra.

\(\Leftrightarrow\hept{\begin{cases}x=y=z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\end{cases}}\Leftrightarrow x=y=z=9\).

Vậy \(minP=\frac{\sqrt{5}}{3}\Leftrightarrow x=y=z=9\).

ko biết

ok bye

* Áp dụng hệ thức 

  \(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\Rightarrow\frac{1}{AC^2}=\frac{1}{AH^2}-\frac{1}{AB^2}\)

\(\Leftrightarrow\frac{1}{AC^2}=\frac{25}{576}-\frac{1}{36}\Leftrightarrow\frac{1}{AC^2}=\frac{1}{64}\Rightarrow AC=8\)cm 

* Áp dụng hệ thực : \(AC^2=HC.BC\)(*)

mà theo Pytago : \(AB^2+AC^2=BC^2\Rightarrow BC^2=36+\frac{576}{25}=\frac{1476}{25}\)

\(\Rightarrow BC=\frac{6\sqrt{41}}{5}\)Thay vào (*) ta được 

\(HC.\frac{6\sqrt{41}}{5}=64\Rightarrow HC\approx8,33\)cm 

\(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\)

\(\Leftrightarrow\frac{1}{1+x^2}+\frac{1}{1+y^2}-\frac{2}{1+xy}\ge0\)

\(\Leftrightarrow\frac{\left(1+y^2\right)\left(1+xy\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}+\frac{\left(1+x^2\right)\left(1+xy\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}-\frac{2\left(1+x^2\right)\left(1+y^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow1+y^2+xy+xy^3+1+x^2+xy+x^3y-2-2x^2-2y^2-2x^2y^2\ge0\)

\(\Leftrightarrow2xy-x^2-y^2+xy^3+x^3y-2x^2y^2\ge0\)

\(\Leftrightarrow\left(xy-1\right)\left(x-y\right)^2\ge0\)

(đúng do \(xy>1,\left(x-y\right)^2\ge0\))

Bất đẳng thức cuối cùng đúng, mà ta biến đổi tương đương nên bất đẳng thức ban đầu cũng đúng. 

Do đó ta có đpcm.