\begin{equation}(a+b)^{2} = a^{2}+2ab+b^{2}\end{equation}
\begin{equation}x=\frac{-b+-\sqrt{b^{2}-4ac}}{2a}\end{equation}
\begin{equation}x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\end{equation}
\begin{equation}= (k+1)n + \frac{k(k+1)}{2}= \frac{(k+1)(2n+k)}{2}\end{equation}
\begin{equation}(2)^{10} = 1024\end{equation}
$\textit{ Italic }$
aashamsakal.......... blogil puthiya post..... PRITHVIRAJINE PRANAYICHA PENKUTTY.......... vayikkane..........
[im]https://sites.google.com/site/classroommaths/hexagon-squ/10.jpeg?attredirects=0&d=1[/im]
$a^{2}$ ഒരു പൂര്ണ്ണവര്ഗ്ഗമാണെങ്കില് $(a\pm(d))^{2 }$ഒരു പൂര്ണ്ണവര്ഗ്ഗമായിരിക്കും.
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9 comments:
\begin{equation}
(a+b)^{2} = a^{2}+2ab+b^{2}
\end{equation}
\begin{equation}
x=\frac{-b+-\sqrt{b^{2}-4ac}}{2a}
\end{equation}
\begin{equation}
x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}
\end{equation}
\begin{equation}
= (k+1)n + \frac{k(k+1)}{2}
= \frac{(k+1)(2n+k)}{2}
\end{equation}
\begin{equation}
(2)^{10} = 1024
\end{equation}
$\textit{ Italic }$
aashamsakal.......... blogil puthiya post..... PRITHVIRAJINE PRANAYICHA PENKUTTY.......... vayikkane..........
[im]https://sites.google.com/site/classroommaths/hexagon-squ/10.jpeg?attredirects=0&d=1[/im]
$a^{2}$ ഒരു പൂര്ണ്ണവര്ഗ്ഗമാണെങ്കില് $(a\pm(d))^{2 }$ഒരു പൂര്ണ്ണവര്ഗ്ഗമായിരിക്കും.
Post a Comment