Đặt OA = a ; OB = b ; OC = c . Khi đó : 

\(OA+OB+OC+AB+BC+AC=a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}\)

AD BĐT Cauchy ta được : \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{a^2+c^2}\ge\sqrt{2}\left(a+b+c\right)\)

Suy ra : l \(\ge\left(\sqrt{2}+1\right)\left(a+b+c\right)\ge\left(\sqrt{2}+1\right)3\sqrt[3]{abc}\)

Có : \(V=V_{OABC}=\dfrac{abc}{6}\)  . Suy ra :   \(l\ge3\left(\sqrt{2}+1\right)\sqrt[3]{6V}\Leftrightarrow V\le\dfrac{l^3}{27\left(\sqrt{2}+1\right)^3.6}=\dfrac{l^3}{162\left(\sqrt{2}+1\right)^3}\)

" = " \(\Leftrightarrow a=b=c\) = \(\dfrac{l\left(\sqrt{2}-1\right)}{3}\)

\(\dfrac{x-1009}{1001}++\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)

\(\Leftrightarrow\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)

\(\Leftrightarrow x-2010=0\Leftrightarrow x=2010\)

Vậy ... 

\(\dfrac{1-2x}{-5}=\dfrac{-5}{1-2x}\left(x\ne\dfrac{1}{2}\right)\) \(\Rightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\) Vậy ...

ĐKXĐ : \(1\le x\le3\)

\(P=\sqrt{x-1}+\sqrt{3-x}\ge\sqrt{3-1}=\sqrt{2}\)  " = " \(\Leftrightarrow\left(x-1\right)\left(3-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

P \(\le\sqrt{\left(1+1\right)\left(x-1+3-x\right)}=2\) ( BĐT B.C.S) " = " \(\Leftrightarrow x-1=3-x\Leftrightarrow x=2\)

Vậy ... 

25% - 2x = 12,5 + 3/4 \(\Leftrightarrow\dfrac{1}{4}-2x=\dfrac{53}{4}\) \(\Leftrightarrow2x=-13\Leftrightarrow x=-\dfrac{13}{2}\)

Gọi số tự nhiên cần tìm là a

Ta có : \(\dfrac{15+a}{38+a}=\dfrac{1}{2}\)  \(\Rightarrow2\left(15+a\right)=38+a\Rightarrow30+2a=38+a\Rightarrow a=8\)

Vậy ... 

Đề bài yêu cầu gì hả bạn ? 

Thấy : \(\sqrt{x^2+x+3}-x^2+1=\sqrt{x^2+x+3}-\left(x^2-1\right)=\dfrac{x^2+x+3-\left(x^2-1\right)^2}{\sqrt{x^2+x+3}+x^2-1}\)

\(=\dfrac{x^2+x+3-x^4+2x^2-1}{...}=\dfrac{-x^4+3x^2+x+2}{...}\)

\(=\dfrac{-\left(x-2\right)\left(x^3+2x^2+x+1\right)}{...}\)

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

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x^2+x+3}-x^2+1}{x^2-4}=\dfrac{-\left(2^3+2.2^2+2+1\right)}{4.\left[\sqrt{2^2+2+3}+2^2-1\right]}=-\dfrac{19}{24}\)

\(S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^n}=\dfrac{1}{1-\dfrac{1}{2}}=2\)

Khi đó : Lim S = Lim 2 = 2