Question

Halle el valor de la incógnita

hola por favor me ayudan es urgente

(1). 50- [3(2x-3)-2(2x-1)] = 23

(2). (x+1)² = x²+3x-5

(3). (2x-1)²+x(x-2) = 5x(x-3)

(4). (2x-1)(3x+2)-(x-2)² = 5x(x-2)+4

(5). (x+2)²+(3x-1)(x+2) = (3x+4)²-5x(x+3)

Answer 1

50- [3(2x-3)-2(2x-1)] = 23

50 - 6x-9 -4x+2 = 23

43-10x = 23

20 = 10x

x = 2

=====

x^2+2x+1 = x^2+3x-5

x^2-x^2+2x-3x = -5-1

-x = -6

x = 6

============

(2x-1)²+x(x-2) = 5x(x-3)

4x^2-4x+1 +x^2-2x = 5x^2-15x

4x^2+x^2-5x^2 -4x-2x+15x = -1

9x = -1

x = -1/9

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(2x-1)(3x+2)-(x-2)² = 5x(x-2)+4

6x^2+x-2 -x^2+4x-4 = 5x^2-10x+4

6x^2-x^2-5x^2 +x+4x+10x = 4+2+4

15x = 10

x = 10/5 = 2/3

============

(x+2)²+(3x-1)(x+2) = (3x+4)²-5x(x+3)

x^2+4x+4 +3x^2+5x-2 = 9x^2+24x+16 -5x^2-15x

x^2+3x^2-9x^2 +4x+5x-24x+15x = 16-4+2

-5x^2= 14

x^2 = -14/5

No tiene solucion (no hay cuadrado de un numero que de negativo)