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ma poate ajuta cineva​

Ma Poate Ajuta Cineva class=

Răspuns :

Punctul b)
[tex] 3x^2 -\dfrac{27}{49} =0 \\ 3x^2 =\dfrac{27}{49} \\ (\sqrt{3} x)^2 =\left(\dfrac{3\sqrt{3}}{7}\right)^2 \\ \sqrt{3} x =\pm \dfrac{3\sqrt{3}}{7} \implies \tt x = \pm \dfrac{3}{7} \\ \\ S=\left\{ -\dfrac{3}{7} , \dfrac{3}{7} \right\}[/tex]

Punctul e)
[tex] \dfrac{x+3}{28}=\dfrac{7}{x+3} \\ (x+3)^2 = 28\cdot 7 \\ (x+3)^2=196 \\ x+3 =\pm 14 \implies x_1= -17, \ x_2 =11 \\ \\ \tt S=\{-17, \ 11 \} [/tex]

b)3x²-27/49=0 ...... 3x²=27/49 | :3 ......... x²=9/49........... x=±rad(9/49)........... x=±3/7 .... => x1=3/7 si x2=-(3/7)

c)x+3/28=7/x+3........ (x+3)²=196 .......... x²+6x+9=196........... x²+6x-187=0 .........

a=1,b=6,c=-187

coeficient=b²-4ac= ........ 6²-4•1•(-187) = ........ 36+748=784

x1,2=-b±rad(coeficient)/2a.... x1,2=-6±rad784/2•1 ......... x1,2=-6±28/2

x1=-6-28/2=-34/2=-17

x2=-6+28/2=22/2=11