Difference between revisions of "Mathematical formula"
From mispar
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[[(3n+1)²=((10·(⅓·3n)²)-(⅓·3n)²)+3n+(3n+1)| | [[(3n+1)²=((10·(⅓·3n)²)-(⅓·3n)²)+3n+(3n+1)| | ||
<math>\scriptstyle\left(3n+1\right)^2=\left[\left[10\sdot\left[\frac{1}{3}\sdot\left(3n\right)\right]^2\right]-\left[\frac{1}{3}\sdot\left(3n\right)\right]^2\right]+3n+\left(3n+1\right)</math>]] | <math>\scriptstyle\left(3n+1\right)^2=\left[\left[10\sdot\left[\frac{1}{3}\sdot\left(3n\right)\right]^2\right]-\left[\frac{1}{3}\sdot\left(3n\right)\right]^2\right]+3n+\left(3n+1\right)</math>]] | ||
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| − | [[category: | + | [[category: #biquadratic equation ]] |
| + | [[comment: 4(x²+8)=x⁴ ]] | ||
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Revision as of 05:46, 5 April 2019
![\scriptstyle\left(3n+1\right)^2=\left[\left[10\sdot\left[\frac{1}{3}\sdot\left(3n\right)\right]^2\right]-\left[\frac{1}{3}\sdot\left(3n\right)\right]^2\right]+3n+\left(3n+1\right)](/mediawiki/images/math/2/a/f/2af90eb131ece9e62879b03d8922a419.png)
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