Cine-mi dă niste idei?
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Emergency stop.
a)Pentru inceput vom egala modulele cu zero;
x+2=0->x=-2 si 3^x-9=0->x=2 . Vom face semnul si vom explicita aceste module
………xl………………………………-2…………………………………2……………
x+2…..l- – – – – – – – – – – – – – –0+ + + + + + + + + + + + ++ + + ++++
3^x-9..l – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – 0+ + + + +
lx+2l=.l………-x-2…………………..l……x+2………………………………………….
l3^x-9l=l………………-3^x+9………………………………………….l..3^x-9….
Ec..l.3^(-x-2)+3^x-9=10.3^x-9…l.3^(x+2)+3^x-9=10.3^x-9..l..3^(x+2)-
………………………………………………………………………………..l.-3^x+9=
………………………………………………………………………………..l.10.3^x-9
solutii..l..1/(9.3^x)=9.3^X->x=-2.l..9.3^x=9.3^x->x=[-2 , 2]…l.3^x=9
………………………………………………………………………………..l->x=2
In final solutia ec este intervalul x=[-2 , +2]
Exercitiul b poti s-l faci si tu ;√(8-2^x)=2-√(2^x-4) si ridici la patrat si este simplu
Exercitiul C)il faci ,explicitand mai intai modulul lx-1l ca la a)