Question

Doar testul 1. Detaliat va rog!

Answer 1

Tot ce se afla sub radical se face conditie de existenta mai mare sau egal decat 0

1 CE x+1 >/ 0 => x>/ -1 , x apatine [-1,infinit)

V(x+1)=x-1 ridici totul la a 2

V(x+1)² = (x-1)²

x+1 = x²-2x+1

x²-2x-x+1-1=0

x²-3x=0

x(x-3)=0

x1=0

x2= 3

2. CE x-1 >/0 , x>/ 1

2-x >/0, -x>/ -2 , x</ 2 => x apartine [1,2]

V(x-1)+V(2-x)=1 ridici totul la a 2

x-1+2V(x-1)(2-x)+2-x = 1

1+2V(x-1)(2-x)=1

2V(x-1)(2-x)=0

(x-1)(2-x)=0

x1=1, x2= 2

3. CE x+1 >/0 , x>/ -1  x apartine [-1,infinit)

x²-x-2=0, x²+x-2x-2=0 , x(x+1)-2(x+1)=0 , (x+1)(x-2)=0 x apartine [-1,2]

Din ambele conditii am rezultat ca x apartine [-1,infinit]

V(x+1)=V(x²-x-2) ridici la a 2 si astfel ai scapat de radical

x+1=x²-x-2

x²-x-x-2-1=0

x²-2x-3=0

x²+x-3x-3=0

x(x+1)-3(x+1)=0

(x+1)(x-3)=0

x1= -1

x2= 3

Bafta!