Question

Sa se rezolve ecuatiile:
a)√2x-1=3
b)∛(2x-1)=3
c) √(2x+3)=4x
d)∛(x+2√2) =√2
e) √(3-x)=-2x
f) √(x^2-3)=3-x
g) ∛(x^3-7)= x-1
h) √x^2-3x+2= 1-x

Answer 1

a) √2x-1=3
conditia de exsistenta 
,2x-1>0,x>1/2
√2x-1=3
2x-1=9
2x=10
x=5

b) ∛(2x-1)=3
se ridica la a 3 a in ambii membrii
2x-1=27
2x=28
x=14

c)√(2x+3)=4x
conditia de existenta
2x+30>0 ;2x>-30:x>-15
ridicand la patrat
2x+3=16x²
16x²-2x-3=0
Δ=4+192=196
x1=(2-14)/8=-12/8=-3/2
x2=(2+14)/8=2

d)ridicand la a 3 a 
[∛(x+2√2)]³ =(√2)³
x+2√2=2√2
x=0

e)√(3-x)=-2x
conditia de existenta
3-x>0 ;x<3
ridicand la patrat
3-x=4x²
4x²+x-3=0
Δ=1+48=49
x1=(-1-7)/8=-1
x2=(-1+7)/8=6/8=3/4

f)√(x^2-3)=3-x
conditia de existenta
x²-3>0 ,x∈(-∞,-√3)∪(√3,+∞)
ridicand la patrat
x²-3=9-6x+x²
6x=12
x=2

g) ∛(x^3-7)= x-1
ridicand la a 3 a
x³-7=x³-3x²+3x-1
3x²-3x-6=0|:3
x²-x-2=0
Δ=1+8=9
x1=(1-3)/2=-1
x2=(1+3)/2=2

h)√x^2-3x+2= 1-x
conditia de exsitenta
x²-3x+2>0
x1=(3-1)/2=1
x2=(3+1)/2=2
x∈(-∞,1)∪(2,+∞)
ridicand la patrat
x²-3x+2=1-2x+x²
x=1