Question

ejercicio 81, segundo punto,álgebra de baldor, ayuda!! es urgente

Answer 1

(3x-7)²-5(2x+1)(x-2)=-x²-[-(3x+1)] 
(3x-7)²-5(2x+1)(x-2)=-x²-[-3x-1]
(3x-7)²-5(2x+1)(x-2)=-x²+3x+1
(3x-7)²-5(2x²-4x+x-2)=-x²+3x+1
(3x-7)²-5(2x²-3x-2)=-x²+3x+1
(3x-7)²-10x²+15x+10=-x²+3x+1
(3x-7)²-10x²+15x+10+x²-3x-1=0
(3x-7)²-9x²+12x+9=0
(3x)²+2(3x)(-7)+(-7)²-9x²+12x+9=0
9x²-42x+49-9x²+12x+9=0
(9x²-9x²)+(-42x+12x)+(+49+9)=0
-30x+58=0
30x=58
x=58/30----x=29/15=1.9333

Comprovación
(3*(29/15)-7)²-5(2*(29/15)+1)({29/15}-2)=-(29/15)²-[-(3*{29/15}+1)] 
Simplificando todo nos da
3.0622=3.0622