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ìܽ

作者: QS   test 2021-04-27 16:38:11  [点击:8389]
(1) \( \) ʾѧʽ(inline math)

\(x \gt 0\), ƽ \(\sqrt{x^2+1}\)...

ʾΪ\(x \gt 0\), ƽ\(\sqrt{x^2+1}\)...


(2) \[ \] ʾѧʽ(block math)

\[ a = \sqrt{1+ \sqrt{1+x^2}}\]

ʾΪ: \[ a = \sqrt{1+ \sqrt{1+x^2}}\]


(3) Latexʹ

: \[\begin{align}
b^2 &= \left(\frac{a}{2} + d \right)^2 + h^2 \\
c^2 &= \left(\frac{a}{2} - d \right)^2 + h^2
\end{align}\]

ʾΪ:
\[\begin{align}
b^2 &= \left(\frac{a}{2} + d \right)^2 + h^2 \\
c^2 &= \left(\frac{a}{2} - d \right)^2 + h^2
\end{align}\]



ʽ
\[
b^2 + c^2 = \frac{a^2}{2} + 2d^2 + 2h^2
\]
ע
\[
L_a^2 = d^2 + h^2
\]

\[
L_a^2 = \frac{1}{2}\left(b^2 + c^2 - \frac{a^2}{2}\right)
\]
\(L_b\)\(b\)ߵߣͬ
\[
L_b^2 = \frac{1}{2}\left(a^2 + c^2 - \frac{b^2}{2}\right)
\]
ڼ \(a \gt b\)
\[
L_a^2 = \frac{1}{2}\left(b^2 + c^2 - \frac{a^2}{2}\right)
\lt \frac{1}{2}\left(a^2 + c^2 - \frac{a^2}{2}\right)
\lt \frac{1}{2}\left(a^2 + c^2 - \frac{b^2}{2}\right)
= L_b^2
\]
\[
L_a \lt L_b
\]
���༭ʱ��: 2021-04-27 18:07:01

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