Question

Va rog ajutatima la evaluare la matematica!.  Deci : Rezolvati in Q euatia: a) x - 1 supra 2 =-1 intreg 1 supra 4;  b) 3x-2= 6; c) 3 inmultit (2x -1) =7 intreg 3 supra 4.     alt exercitiu: 6 intreg 5 supra 8 + 1 intreg 7 supra 12 ; b) - 11 supra 12 + 7 supra 8; Va rog multt imi trebuie acum! ajutatima. 

Answer 1

a) raspunsul este -1 supra 2
b)x=8 supra 3
c)43 supra 4

Answer 2

a.$\frac{x-1}{2}=\frac{-1*4+1}{4}$
$\frac{x-1}{2}=\frac{-3}{4}$ Produsul mezilor si al extremilor (Inmultirea pe diagonala) 
4*(x-1)=-6 => 4x-4=-6 => 4x=-2=> x=-2/4=-1/2 
b.3x-2=6=>3x=8=>x=8/3

c.$7 intreg\frac{3}{4}=\frac{7*4+3}{4}=7 intreg\frac{3}{4}=\frac{31}{4}$
3*(2x-1)=$\frac{31}{4}$
6x-3=$\frac{31}{4}$ | * 4 inmultim toata ecuatia cu 4 pentru a scapa de numitor sau aduci fiecare numar la acelasi numitor...
24x-12=31
24x=43
x=43/24
d.6intreg$\frac{5}{8}$+1intreg $\frac{7}{12}=$
$\frac{6*8+5}{8}+\frac{1*12+7}{12}=$

$\frac{53}{8}+ \frac{19}{12}= \frac{53*3}{8*3}+\frac{19*2}{12*2}=\frac{159}{24}+\frac{38}{24}=\frac{159+38}{24}=\frac{197}{24}$

e.$\frac{-11}{12} + \frac{7}{8}= \frac{-11*2}{12*2} + \frac{7*3}{8*3}= \frac{-22}{24} + \frac{21}{24}= \frac{-22+21}{24}= \frac{-1}{24}$