ITT we confuse the mathsfags

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ITT we confuse the mathsfags

[math] -20 = -20 [/math]

[math] 25 - 45 = 16 - 36 [/math]

[math]5^2 - (5 \times 9) = 4^2 - (4 \times 9) [/math]

[math]5^2 - (5 \times 9) + \frac{81}{4}= 4^2 - (4 \times 9) + \frac{81}{4} [/math]

[math](5-\frac{9}{2})^2 = (4-\frac{9}{2})^2 [/math]

[math]5-\frac{9}{2}= 4-\frac{9}{2} [/math]

[math] 5 = 4 [/math]

checkmate mathfags

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[math] \int tan(x)dx = \int tan(x)dx [/math]

[math] \int tan(x)dx = \int sin(x)sec(x)dx [/math]

[math] \int tan(x)dx = -sec(x)cos(x) + \int cos(x)tan(x)sec(x)dx [/math]

[math] \int tan(x)dx = -1 + \int tan(x) dx [/math]

[math] 0 = -1 [/math]

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Line 5 does not imply line 6

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>>7789777

>comparing indefinite integrals

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[math] x = 0 [/math]

[math] 1 = 0 [/math]

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[math] \frac{1}{3} = 0.\overline{3} [/math]

[math] \frac{2}{3} = 0.\overline{6} [/math]

[math] \frac{3}{3} = 0.\overline{9} [/math]

[math] \frac{3}{3} = 1 = 0.\overline{9} [/math]

eat shit infintesimalfags

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>>7789768

In the transition from line 5 to line 6 of the op, we are supposed to be rused into letting root extraction take place without comment, or consideration as to signing; note that the bases are variously +-0.5.

Since it may happen that (-1)^2 = 1^2, etc, it is not valid to assume that x^2 = y^2 -> x = y. As in, say, the derivation of the quadratic formula, there is a possibility of +- signing, which due to the above requires a bit more investigation.

By quick re-arrangement, one legitimate re-phrase which causes the two to be equal to each other is when the RHS is 9/2 - 4. this forces (correctly), say,

5 = 9 - 4 , or, 5 - 9 = -4.

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