Question

Exercitiul 14 va rogg....multumesc !!

Answer 1

a) (x+1)(x-1)=x^2+5x+4
x^2-1=x^2+5x+4
x^2-1-x^2-5x-4=0
Se reduce x^2 cu -x^2 si ramane:
-1-5x-4=0
-5x-5=0
-5x=5
x=-1
S={-1}

b)(x-2)^2- (x+1)^2=x-4
x^2-4x+4- ( x^2+2x+1)=x-4
x^2-4x+4-x^2-2x-1-x+4=0
Se reduc x^2 cu -x^2 si obtinem:
-4x+4-2x-1-x+4=0
-7x+7=0
-7x=-7
x=1
S={1}

c)(2x+3)^2+7=4(x-2)^2
4x^2+12x+9+7=4(x^2-4x+4)
4x^2+12x+16=4x^2-16x+16
4x^2+12x+16-4x^2+16x-16=0
Se reduce 4x^2 si-4x^2
12x+16+16x-16=0
28x=0
x=0
S={0}

d)(3x-4)^2+20=5(x+1)^2+(2x-3)^2
9x^2-24x+16+20=5(x^2+2x+1) +4x^2-12x+9
9x^2-24x+36=5x^2+10x+5+4x^2-12x+9
9x^2-24x+36=9x^2-2x+14
9x^2-24x+36-9x^2+2x-14=0
Se reduce 9x^2 si -9x^2
-24x+36+2x-14=0
-22x+22=0
-22x=-22
x=1
S={1)

e) (x+2)(3x+2)-3x=2x^2+(x+1)^2
3x^2+2x+6x+4-3x=2x^2+x^2+2x+1
3x^2+5x+4=3x^2+2x+1
3x^2+5x+4-3x^2-2x-1=0
Se reduce 3x^2 cu-3x^2
5x+4-2x-1=0
3x-3=0
3x=3
x=1
S={1}

e)x(x+2)-(2x+3)(3x+4)=18-5x^2
x^2+2x-(6x^2+8x+9x+12)=18-5x^2
x^2+2x-6x^2-17x-12-18+5x^2=0
-5x^2-15x-30+5x^2=0
Se reduce -5x^2 cu 5x^2
-15x-30=0
-15x=30
x=-2
S={-2}