Junior one mathematics super problem! It is known that in △ ABC, high be and CF intersect at O, AE = AF, which means ob = OC

Proband △ Abe ≌ △ ACF,
∴AC=AB,∠ABE=∠ACF,
∵AE=AF,
∴AB-AF=AC-AE,
∴BF=CE,
∵ CF, be intersect o,
∴∠FOB=∠EOC,
≌ syndrome △ FOB ≌ EOC,
∴OB=OC

Calculate the square of 2014x2012 + 1 / 2013

2013^2/(2012*2014+1)=2013^2/((2013+1)(2013-1)+1)=2013^2/2013^2=1

What is the value range of X to minimize the formula | x + 5 | + | X-2 |?

In order to minimize the formula | x + 5 | + | X-2 |, the value range of X is: - 5 ≤ x ≤ 2 Analysis: because the formula is in the form of sum of absolute values, in order to minimize | x + 5 | + | X-2, we should try to minimize each absolute value (equal to 0). When | x + 5 | = 0, x = - 5, | x + 5 | + | X-2 | = 7; when | X-2 | = 0, X

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