144 replies to this topic

[Nesbit]

\begin{align}  
S_{ABCD} & =-R^2\int_{t_1}^{t_2}\frac{dt}{\sqrt{1+\cot^2\varphi _o-t^2}} \\
&=R^2\int_{t_2}^{t_1}\frac{dt}{\sqrt{\frac{1}{\sin^2\varphi _o}-t^2}} \\
&=R^2\left [  \arcsin(t.\sin \varphi _o) \right ] _{\cos \theta _B}^{\cos \theta _A} \\
&=R^2\left [ \arcsin(\sin \varphi _o\cos \theta _A)-\arcsin(\sin \varphi _o\cos \theta _B) \right ] \label{eq:6}
\end{align}

[Nesbit]

Align:

 

\begin{align}  
S_{ABCD} & =-R^2\int_{t_1}^{t_2}\frac{dt}{\sqrt{1+\cot^2\varphi _o-t^2}} \\
&=R^2\int_{t_2}^{t_1}\frac{dt}{\sqrt{\frac{1}{\sin^2\varphi _o}-t^2}} \\
&=R^2\left [  \arcsin(t.\sin \varphi _o) \right ] _{\cos \theta _B}^{\cos \theta _A} \\
&=R^2\left [ \arcsin(\sin \varphi _o\cos \theta _A)-\arcsin(\sin \varphi _o\cos \theta _B) \right ]
\end{align}

 

Equation with boxed:

 

\begin{equation} \label{eq:1} \boxed{\cos \left ( \overrightarrow{OP}, \overrightarrow{OS} \right ) = \frac{\overrightarrow{OP}. \overrightarrow{OS}}{OP.OS} = \cos \varphi_P \cos \varphi_S \cos (\lambda_P - \lambda_S) +\sin \varphi_P\sin \varphi_S} \end{equation}

 

$$\boxed{\cos \left ( \overrightarrow{OP}, \overrightarrow{OS} \right ) = \frac{\overrightarrow{OP}. \overrightarrow{OS}}{OP.OS} = \cos \varphi_P \cos \varphi_S \cos (\lambda_P - \lambda_S) +\sin \varphi_P\sin \varphi_S} $$

[Nesbit]

We have $a^2+b^2 = c^2$.

[PhucLe]

$\sqrt{2}$

[ineX]

\sum \frac{a^{2}b^{2}}{ac+bc} \geq \frac{\sum (ab)^{2}}{2\sum ab} = \frac{1}{2}\sum ab

[ineX]

sao mà nó không ra công thức nhỉ

 

\sum \frac{a^{2}b^{2}}{ac+bc} \geq \frac{\sum (ab)^{2}}{2\sum ab} = \frac{1}{2}\sum ab

[L Lawlist]

$q{1}$

[nozakiboy]

\[\sqrt {{a^2} + {b^2}} \frac{a}{a}{\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}^h}\]

\[{9^2} + 8x = \Delta \]

$x+1=10$

\[\left\{ {_{x - y = 1}^{x + y = 9}} \right.\]

Edited by nozakiboy, 21-02-2016 - 13:30.

[nozakiboy]

\[\sqrt {{a^2} + {b^2}} \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\frac{{n!}}{{r!\left( {n - r} \right)!}}\]

Edited by nozakiboy, 21-02-2016 - 14:04.

[HvuadapchaiH]

$$$fuc(ntion)k$$+$$x\oint_{y}^{z}x$$$

[Zaraki]

$$\begin{aligned} \underset{\equiv 1 \pmod 2 }{ \Large A} & = Y-X \\

\underset{\equiv 1 \pmod 2, \; \le (p-1)/2 }{ \Large A}+\underset{\equiv 1 \pmod 2, \; \ge (p+1)/2 }{ \Large A} & = \left( \underset{\equiv 1 \pmod 2, \; \ge (p+1)/2 }{ \Large A}+ \underset{\equiv 0 \pmod 2, \; \ge (p+1)/2 }{ \Large A} \right)- \left(  \underset{\equiv 1 \pmod 2, \; \le (p-1)/2 }{ \Large A}+ \underset{\equiv 0 \pmod 2, \; \le (p-1)/2 }{ \Large A} \right) \\

2 \times \underset{\equiv 1 \pmod 2, \; \le (p-1)/2 }{ \Large A} & = \underset{\equiv 0 \pmod 2, \; \ge (p+1)/2 }{ \Large A} - \underset{\equiv 0 \pmod 2, \; \le (p-1)/2 }{ \Large A} \\

 \underset{\equiv 2 \pmod 4 }{ \Large A} & = \underset{\equiv 0 \pmod 2, \; \ge (p+1)/2 }{ \Large A} - \underset{\equiv 0 \pmod 2, \; \le (p-1)/2 }{ \Large A} \\ 

 \underset{\equiv 2 \pmod 4, \le (p-1)/2 }{ \Large A}+  \underset{\equiv 2 \pmod 4, \ge (p+1)/2}{ \Large A} & = \left(  \underset{\equiv 2 \pmod 4, \ge (p+1)/2}{ \Large A}+  \underset{\equiv 0 \pmod 4, \le (p-1)/2}{ \Large A} \right) -  \underset{\equiv 0 \pmod 2, \le (p-1)/2 }{ \Large A} \\

 \underset{\equiv 2 \pmod 4, \le (p-1)/2}{ \Large A} &=  \underset{\equiv 0 \pmod 4, \ge (p+1)/2}{ \Large A} -  \underset{\equiv 0 \pmod 2, \le (p-1)/2 }{ \Large A} \\

 \underset{\equiv 2 \pmod 4, \le (p-1)/2}{ \Large A} & = \left(  \underset{\equiv 4 \pmod 8, \ge (p+1)/2 }{ \Large A} +  \underset{\equiv 0 \pmod 8, \ge (p+1)/2}{ \Large A} \right) - \left(  \underset{\equiv 2 \pmod 4, \le (p-1)/2}{ \Large A}+  \underset{\equiv 0 \pmod 4, \le(p-1)/2}{ \Large A} \right) \\

 \underset{\equiv 2 \pmod 4, \ge (p-3)/4, \le (p-1)/2}{ \Large A} +  \underset{\equiv 2 \pmod 4, \le(p-3)/4}{ \Large A} & = 2 \left(  \underset{\equiv 2 \pmod 4, \ge (p-3)/4, \le (p-1)/2}{ \Large A}+  \underset{\equiv 0 \pmod 4, \ge (p-3)/4, \le (p-1)/2}{ \Large A} \right) - \left(  \underset{\equiv 2 \pmod 4, \le (p-1)/2}{ \Large A}+  \underset{\equiv 0 \pmod 4, \le(p-1)/2}{ \Large A} \right) \\

\underset{\equiv 2 \pmod 4, \ge (p-3)/4, \le (p-1)/2}{ \Large A} +  \underset{\equiv 2 \pmod 4, \le(p-3)/4}{ \Large A} & = \left(  \underset{\equiv 2 \pmod 4, \ge (p-3)/4, \le (p-1)/2}{ \Large A}+  \underset{\equiv 0 \pmod 4, \ge (p-3)/4, \le (p-1)/2}{ \Large A} \right) - \left(  \underset{\equiv 2 \pmod 4, \le (p-3)/4}{ \Large A}+  \underset{\equiv 0 \pmod 4, \le(p-3)/4}{ \Large A} \right) \\

2 \times \underset{\equiv 2 \pmod 4, \le (p-3)/4}{ \Large A} & = \underset{\equiv 0 \pmod 4, \ge (p-3)/4, \le (p-1)/2}{ \Large A} - \underset{\equiv 0 \pmod 4, \le (p-3)/4}{ \Large A} \\ 

\underset{\equiv 4 \pmod 8, \le (p-1)/2}{ \Large A} & = \underset{\equiv 0 \pmod 4, \ge (p-3)/4, \le (p-1)/2}{ \Large A} - \underset{\equiv 0 \pmod 4, \le (p-3)/4}{ \Large A} \\

\underset{\equiv 4 \pmod 8, \ge (p-3)/4, \le (p-1)/2}{ \Large A} + \underset{\equiv 4 \pmod 8, \le (p-3)/4}{ \Large A} & = \underset{\equiv 0 \pmod 4, \ge (p-3)/4, \le (p-1)/2}{ \Large A} - \underset{\equiv 0 \pmod 4, \le (p-3)/4}{ \Large A} \\

\underset{\equiv 4 \pmod 8, \le (p-3)/4}{ \Large A} + \underset{\equiv 0 \pmod 4, \le (p-3)/4}{ \Large A} & = \underset{\equiv 0 \pmod 8, \ge (p-3)/4, \le (p-1)/2}{ \Large A} \\ 

2 \times \underset{\equiv 4 \pmod 8, \le (p-3)/4}{ \Large A} + \underset{\equiv 0 \pmod 8, \le (p-3)/4}{ \Large A} & = \underset{\equiv 0 \pmod 8, \ge (p-3)/4, \le (p-1)/2}{ \Large A} \\

2 \left( \times \underset{\equiv 4 \pmod 8, \le (p-3)/4}{ \Large A} + \underset{\equiv 0 \pmod 8, \le (p-3)/4}{ \Large A} \right) & =  \underset{\equiv 0 \pmod 8, \le (p-3)/4}{ \Large A} +\underset{\equiv 0 \pmod 8, \ge (p-3)/4, \le (p-1)/2}{ \Large A} \\ 

2 \times \underset{\equiv 0 \pmod 4, \le (p-3)/4}{ \Large A} & = \underset{\equiv 0 \pmod 8 \le (p-1)/2}{ \Large A}

\end{aligned}$$ 

[thaomy1106]

1.Cho đa thức P(x) = ax² + bx +c đồng thời thỏa mãn P(x) $\geq$ 0 với mọi x thuộc R và b>a

Tìm Min P = $\frac{a+b+c}{b-a}$

 

[tanthanh112001]

$\left\{\begin{matrix} &x+y=0 & \\ &3x-4y=0 & \end{matrix}\right.$

[tanthanh112001]

$ \stackrel\frown{AB}$

[thanhmylam]

$$\begin{array}{|c|c|c|}

 

\hline

\end{array}$$

[tanthanh112001]

$x^{2}$, $x_{3}$

Edited by tanthanh112001, 25-03-2016 - 12:17.

[adamfu]

$/sqrt$$/fract$

[vamath16]

$\doteq$

[vamath16]

$\fn_jvn n \to$

[vamath16]

<a href="http://www.codecogs....;&space;a_4}}}"target="_blank"><img src="http://latex.codecogs.com/gif.latex?x&space;=&space;a_0&space;&plus;&space;\frac{1}x&space;=&space;a_0&space;&plus;&space;\frac{1}{\displaystyle&space;a_1&space;&plus;&space;\frac{1}{\displaystyle&space;a_2&space;&plus;&space;\frac{1}{\displaystyle&space;a_3&space;&plus;&space;a_4}}}{a_1&space;&plus;&space;\frac{1}{a_2&space;&plus;&space;\frac{1}{a_3&space;&plus;&space;a_4}}}" title="x = a_0 + \frac{1}x = a_0 + \frac{1}{\displaystyle a_1 + \frac{1}{\displaystyle a_2 + \frac{1}{\displaystyle a_3 + a_4}}}{a_1 + \frac{1}{a_2 + \frac{1}{a_3 + a_4}}}" /></a>