pratikum 1 hardiansyah

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x 1,2 = b 2 4 ac 2 a A = [ 1 2 3 4 ] 1 3 ( x 2 +5 x +6 ) dx f ( x )= x+ log x 2 2 x 2 + 1

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Page 1: Pratikum 1 hardiansyah

x1,2=−b±√b2−4ac

2a

A=[1 23 4 ]

∫1

3

(x2+5 x+6 )dx

f ( x )= x+ log x2❑2

x2+1