Răspunsuri

2014-04-26T17:07:51+03:00

4(2x+1)²-(3x+1)²=0⇒(8x+4)²=(3x+1)²⇒64x²+64x+16=9x²+6x+1⇒55x²+58x+15=0

Δ=58²-4*55*15=3364-3300=64

x(1)=(-58+√64)/110⇒x(1)=(-58+8)/110⇒x(1)=-50/110⇒x(1)=-5/11

x(2)=(-58-√64)/110⇒x(2)=(-58-8)/110⇒x(2)=-64/110⇒x(2)=-32/55

2014-04-26T17:32:23+03:00
4(2x +1)²-(3x+1)²=0⇒(8x+4)²=(3x+1)²⇒64x²+64x+16=9x²+6x+1⇒55x²+58x+15=0Δ=58²-4*55*15=3364-3300=64x(1)=(-58+√64)/110⇒x(1)=(-58+8)/110⇒x(1)=-50/110⇒x(1)=-5/11x(2)=(-58-√ 64)/110⇒x(2)=(-58-8)/110⇒x(2)=-64/110⇒x(2)=-32/55(ma bucur daca te-am ajutat  :* )