Waste less time on Facebook — follow Brilliant.
×

Ashish's \(\LaTeX\) playground

\(\large \text {This is my note for testing} \LaTeX \) \(:P\)

\(\large \text {A special thanks to}\) \(\large \color {red}{\text {Pi Han Goh}}\) \(\large \text {and}\) \(\large \color {blue}{\text {Andrew Ellinor}}\) \(\large \text {and}\) \(\large \color {green}{\text {Nihar Mahajan}}\)\(\large \text {for the success of this note.}\) \[\begin {align} \begin{array} x & = 2+5 \\ & = 7 \end {array} \end {align}\]

\(\begin {pmatrix} 1 & & 0\\ 0 & & 1\ \end {pmatrix}\) × \(\begin {pmatrix} 1 & & 0\\ 0 & & 1\ \end {pmatrix}\) = \(\begin {pmatrix} 1+0 & & 0+0\\ 0+0 & & 0+1\ \end {pmatrix}\) = \(\begin {pmatrix} 1 & & 0\\ 0 & & 1\ \end {pmatrix}\)

\[\large{\color{magenta}{\text{My name is Ashish}}}\]


\[\begin{align} & x + 7x & = 24\\ \implies & 8x & =24\\ \implies & x & = \dfrac {24}{8}\\ \therefore & x & = \boxed{3} \end{align}\]


\[\ce {{Na}_2CO_3 + H_2O -> NaOH + H_2O + CO_2}\]


\[\begin{pmatrix} \nearrow \circ \nwarrow & \approx & \nearrow \circ \nwarrow\\ & \bigcup & \\ & \nearrow \searrow \nearrow \searrow \nearrow \searrow & \\& \smile & \end{pmatrix}\]


\[ \LARGE{ \require{enclose} \begin{array}{rll} \phantom{0}\ \boxed{{5}} && \\[-2pt] \boxed{{5}}\ \enclose{longdiv}{\boxed{{3}} \ \boxed{7}}\kern-.2ex \\[-2pt] \underline{\boxed{{3}} \ \boxed{{5}}} && \\[-2pt] \boxed{{2}} \end{array} } \]


\[ \LARGE{ \require{enclose} \begin{array}{rll} \boxed{{x}}\ + \boxed{{1}} && \\[-2pt] \boxed{{x}}\ + \boxed{{1}} \enclose{longdiv}{\boxed{{x^2}}\ + \boxed{{2x}}\ + \boxed{{1}}}\kern-.2ex \\[-2pt] \underline{\boxed{{x^2}}\ + \boxed{{x}}\ \phantom{0}\ \phantom{0}\ \phantom{0}} \phantom{0}\\[-2pt] \phantom{0}\ \boxed{{x}}\ + \boxed{{1}}\kern-.2ex \\[-2pt] \underline{\boxed{{x}}\ + \boxed{{1}}} && \\[-2pt] \phantom{0}\ \phantom{0}\ \boxed{{0}} \end{array} } \]


\[\begin {align} \begin {array} x & \dfrac{29 \times 7}{116}\\ = & \dfrac{{\require{cancel}{\cancel {29}}} \times 7}{{\require {cancel}{\cancel {116}}} 4}\\ = & \boxed {\dfrac{7}{4}} \end {array} \end {align}\]


\[\huge \text{I ♡ Brilliant}\]


\[\large \text{Here are some symbols} :- ♤■□●◇◆•○♡♧¡¿《》¤°♧▪\]


\[\begin {eqnarray} \dfrac {10}{20} & =\require {cancel}{\dfrac {\xcancel{10}}{\xcancel{20}2}}\\ & = \dfrac {1}{2} \end {eqnarray}\]


\[{\Huge {\color {red}{B}}}{\huge{\color {blue}{R}}}{\LARGE {\color {yellow}{I}}}{\Large{\color{green}{L}}}{\large{\color {magenta}{L}}}{\Large{\color{purple}{I}}}{\LARGE{\color{black}{A}}}{\huge{\color {pink}{N}}}{\Huge{\color{brown}{T}}}\]


\[\begin{align} 5+4 & = 8 +1\\ & = 9 \\ \\ 7+2 & = 8+1\\ & = 9 \end{align}\]


\[\Huge \sqrt[\sqrt[x^5]{\sqrt[x^4]{\sqrt[x^3]{\sqrt[x^2]{\sqrt[x]{x}}}}}]{\sqrt[5\sqrt[x]{\sqrt[2x]{\sqrt[3x]{\sqrt[4x]{x}}}}]{\sqrt[5x]{\sqrt[4x]{\sqrt[3x]{\sqrt[2x]{x}}}}}} = x^{x^{5 + \tfrac{x^{11} + 24}{24x^{15}}}}\]

Find the real value of \(x\) satisfying the real equation above.

\(\text{Note}\):- Here \(x \neq \{-1 , 0 , 1\}\)


\(\sqrt[\sqrt[4]{{\left(\sqrt[x]{4x^4}\right)}^{x^4}}]{\sqrt[4]{\sqrt[x]{4}}}\) = \(4^{-17x^{-x^3 - 1}}\)

\(\sqrt[4]{{\left(\sqrt[x]{4x^4}\right)}^{x^4}}\)

\({\sqrt[4]{\sqrt[x]{4}}}^{\tfrac{1}{\sqrt[4]{{\left(\sqrt[x]{4x^4}\right)}^{x^4}}}}\)

\(\log_8 \log_4 \color{blue}{\log_2 16}\\ = \log_8 \log_4 \color{blue}{4}\\ = \log_8 \color{red}{\log_4 4}\\ = \log_8 \color{red}{1}\\ = \color{green}{\log_8 1}\\ = \color{purple}{\boxed{0}}\)

\(\log_{\color{blue}{b}} \color{red}{a} = \dfrac{\log \color{red}{a}}{\log \color{blue}{b}}\)

\(\log_{\color{blue}{\sqrt{2}}} \color{red}{m}\\ \\ = \dfrac{\log \color{red}{m}}{\log \color{blue}{2^{\tfrac{1}{2}}}}\\ \\ = \dfrac{\log m}{\log 2^{\color{green}{\tfrac{1}{2}}}}\\ \\ = \dfrac{\log m}{\color{green}{\dfrac{1}{2}} \log 2}\\ \\ = \dfrac{1}{\frac{1}{2}} \dfrac{\log \color{cyan}{m}}{\log \color{magenta}{2}}\\ \\ = \boxed{2} \log_{\color{magenta}{2}} \color{cyan}{m}\)

Note by Ashish Siva
8 months, 1 week ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

THANK YOU Ashish for creating this note ..helped me to learn latex..thanks a ton.!!:) Rishabh Tiwari · 6 months, 3 weeks ago

Log in to reply

@Rishabh Tiwari Entirely welcome Ashish Siva · 6 months, 3 weeks ago

Log in to reply

Any suggestions? I am practicing this for mastering LaTeX especially long division of polynomials. Ashish Siva · 8 months, 1 week ago

Log in to reply

Log in to reply

@Pi Han Goh Thank u sir. Im a beginner now I can learn latex. Rishabh Tiwari · 6 months, 3 weeks ago

Log in to reply

@Pi Han Goh Thank you, accept my THANKS, ;P Ashish Siva · 8 months, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...