Respuesta:
Explicación paso a paso:
[tex]\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[/tex]
Paso 1
[tex]\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[/tex]
Paso 2
[tex]2x^2-7x+4=2x^2+2x-32[/tex]
Paso 3
[tex]2x^2-7x=2x^2+2x-36[/tex]
Paso 4
[tex]-9x=-36[/tex]
Paso 5
[tex]x=4[/tex]
[tex]\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}[/tex]
[tex]\left(3x-1\right)^2-5\left(x-2\right)-\left(2x+3\right)^2-\left(5x+2\right)\left(x-1\right)=0[/tex]
[tex]-20x+4=0[/tex]
[tex]x=\frac{1}{5}[/tex]
Bueno Te Hice 2 Ejercicios. Solo remplazar y Hallar x.. Tengo 12
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Verified answer
Respuesta:
Explicación paso a paso:
[tex]\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[/tex]
Paso 1
[tex]\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[/tex]
Paso 2
[tex]2x^2-7x+4=2x^2+2x-32[/tex]
Paso 3
[tex]2x^2-7x=2x^2+2x-36[/tex]
Paso 4
[tex]-9x=-36[/tex]
Paso 5
[tex]x=4[/tex]
[tex]\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}[/tex]
Paso 1
[tex]\left(3x-1\right)^2-5\left(x-2\right)-\left(2x+3\right)^2-\left(5x+2\right)\left(x-1\right)=0[/tex]
Paso 2
[tex]-20x+4=0[/tex]
Paso 3
[tex]x=\frac{1}{5}[/tex]
Bueno Te Hice 2 Ejercicios. Solo remplazar y Hallar x.. Tengo 12