Question

(3x-1)²=(2x+3)²+(x+2)² ; x=?

Answer 1

9x²-6x+1=4x²+12x+9+x²+4x+4
9x²-6x+1=5x²+16x+13
9x²-5x²-6x-16x+1-13=0
4x²-22x-12=0  l:2
2x²-11x-6=0
Δ=121+48
Δ=169 =>√Δ=13
x₁=(11+13)/4=6
x₂=(11-13)/4=-1/2

Answer 2

(3x-1)^2=(2x+3)^2+(x+2)^2
9x^2-6x+1=4x^2+12x+9+x^2+4x+4
9x^2-6x+1=5x^2+16x+13
9x^2-6x+1-5x^2-16x-13=0
4x^2-22x-12=0
2(2x-11x-6)=0
2(2x^2+x-12x-6)=0
2(x•(2x+1)-6(2x+1))=0
2(x-6)•(2x+1)=0
(x-6)•(2x+1)=0

x-6=0
2x+1=0

x=6
x=- 1/2

x indice 1=- 1/2
x indice 2=6

^ înseamnă la puterea
cum se face indicele ai în poza



Succes!!!




➡ArianaEdaline