f(x)  \(=2\left(x+3\right)+3\left(x-1\right)\)  = 5x + 3

f(x) = 0 \(\Leftrightarrow5x+3=0\Leftrightarrow x=-\dfrac{3}{5}\)

 B . was injured ( thể bị động )  

Chọn D 13021 

\(BC=2R=\dfrac{BC}{sin\widehat{BAC}}\Rightarrow\widehat{BAC}=90^o\) 

\(\dfrac{3}{20}+\dfrac{1}{6}=\dfrac{9}{60}+\dfrac{10}{60}=\dfrac{19}{60}\)

\(\alpha;\beta\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow cos\alpha;sin\beta>0\)

Ta có : \(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\dfrac{1}{5}}=\dfrac{2}{\sqrt{5}}\)  

\(sin\beta=\sqrt{1-cos^2B}=\sqrt{1-\dfrac{1}{10}}=\dfrac{3}{\sqrt{10}}\)

\(cos\left(\alpha+\beta\right)=cos\alpha.cos\beta-sin\alpha.sin\beta\) = \(\dfrac{2}{\sqrt{5}}.\dfrac{1}{\sqrt{10}}-\dfrac{1}{\sqrt{5}}.\dfrac{3}{\sqrt{10}}=\dfrac{-\sqrt{2}}{10}\)

Số 92 

Gọi cd ; cr của khu vườn lần lượt là : a ; b ( a > b > 0 ) 

Ta có : \(\left\{{}\begin{matrix}a-b=10\\2.2\left(a+b\right)=560\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=10\\a+b=140\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=75\\b=65\end{matrix}\right.\)  (m)

Vậy ... 

P/s : Mik nghĩ là \(\left(2x+1\right)^2\)

\(C=x+\dfrac{1}{4x}+\dfrac{x}{\left(2x+1\right)^2}=\left[\dfrac{x}{\left(2x+1\right)^2}+\dfrac{2x+1}{16}+\dfrac{2x+1}{16}+\dfrac{1}{16x}\right]+\dfrac{3}{4}\left(x+\dfrac{1}{4x}\right)-\dfrac{1}{8}\)

AD BĐT AM - GM ta được : \(\dfrac{x}{\left(2x+1\right)^2}+\dfrac{2x+1}{16}+\dfrac{2x+1}{16}+\dfrac{1}{16x}\ge4\sqrt[4]{\dfrac{1}{16^3}}=\dfrac{1}{2}\)

\(x+\dfrac{1}{4x}\ge2\sqrt{\dfrac{1}{4}}=1\) 

Suy ra : \(C\ge\dfrac{1}{2}+\dfrac{3}{4}.1-\dfrac{1}{8}=\dfrac{9}{8}\)

" = " \(\Leftrightarrow x=\dfrac{1}{2}\)