Question

●(x-2)²+x(x-3)=3(x+4)(x-3)-(x+2)(x-1)+2.
●(3x-1)²-5(x-2)-(2x+3)²-(5x+2)(x-1)=0.
● 2(x-3)-3(x+1)+(x-5)(x-3)+4x*–5x+1)=4x²-12. ● 5(x-2)²-5(x+3)2+(2x-1)(x+2) - 10x =0. ●x²-5x+15=x(x-3)-14+5(x-2)+3(13-2x).
●3(5x-6)(3x+2)-6(3x+4)(x-1)-3(9x+1)(x-2)=0.
●7(x-4)²-3(x+5)²=4(x+1)(x-1)-2.
●5(1-x)²-6(x²-3x-7)=x(x-3)- 2x(x+5)-2​

Answer 1

Respuesta:

• x=4

• x= 1/5
• x=0
• x= 2/7  
• x=1/2
• x=  -7/3

Explicación paso a paso:

Answer 2

$\left(x-2\right)^2+x\left(x-3\right)=3\left(x+4\right)\left(x-3\right)-\left(x+2\right)\left(x-1\right)+2\quad :\quad x=4$

Paso 1

$\left(x-2\right)^2+x\left(x-3\right)=3\left(x+4\right)\left(x-3\right)-\left(x+2\right)\left(x-1\right)+2$

Paso 2

$2x^2-7x+4=2x^2+2x-32$

Paso 3

$2x^2-7x=2x^2+2x-36$

Paso 4

$-9x=-36$

Paso 5

$x=4$

$\left(3x-1\right)^2-5\left(x-2\right)-\left(2x+3\right)^2-\left(5x+2\right)\left(x-1\right)=0\quad :\quad x=\frac{1}{5}$

Paso 1

$\left(3x-1\right)^2-5\left(x-2\right)-\left(2x+3\right)^2-\left(5x+2\right)\left(x-1\right)=0$

Paso 2

$-20x+4=0$

Paso 3

$x=\frac{1}{5}$

Bueno Te Hice 2 Ejercicios. Solo remplazar y Hallar x.. Tengo 12