Test công thức toán

Nắng

Anh sẽ vì em làm cha thằng bé

• 30/12/13

• #201

${\sin }^{4}x+{\cos }^{4}x={({\sin }^{2}x+{\cos }^{2}x)}^2-2\text{s}i{n}^{2}x.{\cos }^{2}x=1-{\displaystyle \frac{{(2\text{s}inx\cos x)}^2}{2}}=1-{\displaystyle \frac{{\sin }^{2}2x}{2}}<br{\displaystyle \frac{}{}}>{\int }_0^{{\displaystyle \frac{\pi }{2}}}{\cos }_{2}x({\sin }^{4}x+{\cos }^{4}x)dx={\int }_0^{{\displaystyle \frac{\pi }{2}}}{\cos }_{2}x(1-{\displaystyle \frac{{\sin }^{2}2x}{2}})dx={\int }_0^{{\displaystyle \frac{\pi }{2}}}{\cos }_{2}xdx-{\displaystyle \frac{1}{4}}{\int }_0^{{\displaystyle \frac{\pi }{2}}}{\sin }^{2}2xd(\sin 2x)<br{\displaystyle \frac{}{}}>$

 

Nắng

Anh sẽ vì em làm cha thằng bé

• 30/12/13

• #202

Lil.Tee a có cách nào khắc phục lỗi công thức dài quá nó không hiển thị hết không :D
Như ở đây này a :D http://olympiavn.org/forum/index.php?topic=42381.15

 

Nắng

Anh sẽ vì em làm cha thằng bé

• 30/12/13

• #203

$\int x\tan xdx<br{\displaystyle \frac{}{}}>\int x\cot xdx<br{\displaystyle \frac{}{}}>\int {\displaystyle \frac{\sin x}{x}}dx<br{\displaystyle \frac{}{}}>\int {\displaystyle \frac{\cos x}{x}}dx<br{\displaystyle \frac{}{}}>\int e^{x^2}dx<br{\displaystyle \frac{}{}}>\int e^{x}lnxdx<br{\displaystyle \frac{}{}}>\int \sqrt{lnx}dx<br{\displaystyle \frac{}{}}>\int {\displaystyle \frac{dx}{lnx}}<br{\displaystyle \frac{}{}}>\int ln(\cos x)dx<br{\displaystyle \frac{}{}}>\int ln(\sin x)dx<br{\displaystyle \frac{}{}}>\int \cos \left(x^2\right)dx<br{\displaystyle \frac{}{}}>\int \sin \left(x^2\right)dx<br{\displaystyle \frac{}{}}>{\int }_{}^{}{e}^{\sin x}dx<br{\displaystyle \frac{}{}}>{\int }_{}^{}{e}^{\cos x}dx<br{\displaystyle \frac{}{}}>{\int }_{}^{}{e}^{\tan x}dx<br{\displaystyle \frac{}{}}>{\int }_{}^{}{e}^{\cot x}dx<br{\displaystyle \frac{}{}}>{\int }_{}^{}\sqrt{x^3+1}dx<br{\displaystyle \frac{}{}}>{\int }_{}^{}{\displaystyle \frac{dx}{(e^x+1)(x^2+1)}}<br{\displaystyle \frac{}{}}>{\int }_{}^{}\sqrt{\cos x}dx<br{\displaystyle \frac{}{}}>{\int }_{}^{}\sqrt{\sin x}dx$

 

Nắng

Anh sẽ vì em làm cha thằng bé

• 30/12/13

• #204

${\int }_{-1}^{1}ln{\displaystyle \frac{{(1+e^x)}^{1+e^{-x}}}{1+e^{-x}}}dx$

 

Tăng Hải Tuân

Well-Known Member

Administrator

• 30/12/13

• #205

s2_la đã viết:

Lil.Tee a có cách nào khắc phục lỗi công thức dài quá nó không hiển thị hết không :D
Như ở đây này a :D http://olympiavn.org/forum/index.php?topic=42381.15

Click để xem thêm...

Cái này đơn giản thôi, nhưng nó không đẹp.

 

NTH 52

Bùi Đình Hiếu

Super Moderator

• 10/1/14

• #206

$$3\sqrt[9] {\dfrac{9\sqrt{3}\left(\sqrt{x^3+4}+\sqrt{3}\right)}{14+2x^3+2x^2-2x+4\sqrt{\left(x^2-2x\right)\left(x^3+4\right)}+4\sqrt{3x^2-6x}+4\sqrt{3x^3+12}}}+\sqrt[3] {\dfrac{6\sqrt{\left(x^2-2x\right)\left(x^3+4\right)}{\left(\sqrt{x^3+4}+\sqrt{3}\right)\left(\sqrt{3}\sqrt{x^3+4}+\sqrt{3}\right)}}=|1-2x|+|2x+3|.$$

 

NTH 52

Bùi Đình Hiếu

Super Moderator

• 16/1/14

• #207

${\displaystyle \frac{1}{\sqrt{1+{(\cot \phi -{\displaystyle \frac{1}{\sin \phi }})}^2}}}$

 

NTH 52

Bùi Đình Hiếu

Super Moderator

• 16/1/14

• #208

$$U_{AN}+U_{NB}={\displaystyle \frac{U}{\cos {\displaystyle \frac{\pi }{5}}}}[\sin ({\displaystyle \frac{\pi }{2}}-\phi )+\sin ({\displaystyle \frac{\pi }{5}}-\phi )].$$

$$={\displaystyle \frac{2U}{\cos {\displaystyle \frac{\pi }{5}}}}\sin {\displaystyle \frac{3\pi }{10}}\cos ({\displaystyle \frac{7\pi }{20}}-\phi )\le {\displaystyle \frac{2U}{\cos {\displaystyle \frac{\pi }{5}}}}\sin {\displaystyle \frac{3\pi }{10}}.$$

 

NTH 52

Bùi Đình Hiếu

Super Moderator

• 16/1/14

• #209

$$U_{AN}+U_{NB}={\displaystyle \frac{U}{\cos {\displaystyle \frac{\pi }{5}}}}(\sin ({\displaystyle \frac{\pi }{2}}-\phi )+\sin ({\displaystyle \frac{\pi }{5}}+\phi )).$$

$$=2{\displaystyle \frac{U}{\cos {\displaystyle \frac{\pi }{5}}}}\sin \left({\displaystyle \frac{7\pi }{20}}\right)\cos ({\displaystyle \frac{3\pi }{20}}-\phi ).$$

$$\le 2{\displaystyle \frac{U}{\cos {\displaystyle \frac{\pi }{5}}}}\sin {\displaystyle \frac{7\pi }{20}}.$$

 

NTH 52

Bùi Đình Hiếu

Super Moderator

• 19/1/14

• #210

$$(\overrightarrow{U},\overrightarrow{U_{MB}})={\displaystyle \frac{\pi }{3}}.$$

 

minhdoan

New Member

• 25/1/14

• #211

${\displaystyle \frac{a}{b}}$

 

minhdoan

New Member

• 25/1/14

• #212

${}_{92}^{234}U$

 

G

guest-1645204321

Guest

• 1/2/14

• #213

$ab$

 

Tăng Hải Tuân

Well-Known Member

Administrator

• 19/3/14

• #214

${\displaystyle \frac{\sqrt{a^2+b^2}}{{\displaystyle \frac{c}{a+{\displaystyle \frac{b}{c}}}}}}$

 

hoangnhatphongt32

Member

• 30/10/14

• #215

$\sqrt{x^2}$ $\alpha$ $\omega$

 

hoangnhatphongt32

Member

• 30/10/14

• #216

${\displaystyle \frac{x}{y}}$ $\to$ $A^4$ $\phi$ $\delta$

 

Last edited: 30/10/14

N

NMH

New Member

• 11/11/14

• #217

${\displaystyle \frac{5T}{6}}$

 

NTH 52

Bùi Đình Hiếu

Super Moderator

• 31/12/14

• #218

$\{\begin{array}{l}x\\ y\\ z\end{array}$

 

NTH 52

Bùi Đình Hiếu

Super Moderator

• 3/1/15

• #219

$\text{arc}\sin (-{\displaystyle \frac{1}{\sqrt{5}}}right)$

 

THL

New Member

• 18/6/15

• #220

$\sqrt{x^2+y^2}-{\displaystyle \frac{R^{2}C}{2L}}+3\pi$