Question

DAU COROANA HAI CA-S USOARE PT VOI !!
a) (x-2)^2-(x+1)^2=x-4
b)(2x+3)^2+7=4(x-2)^2
c)(3x-4)^2+29=5(x+1)^2+(2x-3)^2
d)(x+2)(3x+2)-3x=2x^2+(x+1)^2
e)x(x+2)-(2x+3)(3x+4)=18-5x

Answer 1

a) (x-2)^2-(x+1)^2=x-4
 (x-2-x-1)(x-2+x+1)=x-4
-3(2x-1)=x-4
-6x+3=x-4
7x=7
x=1

b)(2x+3)^2+7=4(x-2)^2
(2x+3)^2-4(x-2)^2=-7
[2x+3-2(x-2)][2x+3+2(x-2)]=-7
(2x+3-2x+4)(2x+3+2x-4)=-7
7(4x-1)=-7
4x-1=-1
4x=0
x=0

c)(3x-4)^2+29=5(x+1)^2+(2x-3)^2
9x²-24x+16+29=5x²+10x+5+4x²-12x+9
-24x+45=-2x+14
22x=31
x=31/22

d)(x+2)(3x+2)-3x=2x^2+(x+1)^2
3x^2+6x+2x+4-3x=2x^2+x^2+2x+1
3x^2+5x+4=3x^2+2x+1
3x+4=1
3x=-3
x=-1

e)x(x+2)-(2x+3)(3x+4)=18-5x
x^2+2x-6x^2-9x-8x-12=18-5x
-5x^2-15x-12=18-5x
-5x^2-10x-30=0
x^2+2x+6=0   
Δ=b²-4ac=2²-4*1*6=4-24=-20    ⇒ ecuatia nu are radacini reale
i^2=-1
x₁=(-2-√-20)/2=-1-i√5
x₁=(-2+√-20)/2=-1+i√5