Question

F(x)=(3x²-1)²+2÷(2x-1)²

Answer 1

Respuesta:

$f(x)=\left(3x^2-1\right)^2+2\div \left(2x-1\right)^2=\left(3x^2-1\right)^2+\frac{2}{\left(2x-1\right)^2}=\frac{2}{\left(2x-1\right)^2}+\frac{\left(3x^2-1\right)^2\left(2x-1\right)^2}{\left(2x-1\right)^2}=\frac{2+\left(3x^2-1\right)^2\left(2x-1\right)^2}{\left(2x-1\right)^2};$

Operamos independientemente el numerador:

$2+\left(3x^2-1\right)^2\left(2x-1\right)^2=\\=2+\left(9x^4-6x^2+1\right)\left(2x-1\right)^2=\\=2+\left(9x^4-6x^2+1\right)\left(4x^2-4x+1\right)=\\=2+9x^4\cdot \:4x^2+9x^4\left(-4x\right)+9x^4\cdot \:1+\left(-6x^2\right)\cdot \:4x^2+\left(-6x^2\right)\left(-4x\right)+\left(-6x^2\right)\cdot \:1+1\cdot \:4x^2+1\cdot \left(-4x\right)+1\cdot \:1=\\=2+9\cdot \:4x^4x^2-9\cdot \:4x^4x+9x^4-6\cdot \:4x^2x^2+6\cdot \:4x^2x-6x^2+4x^2-4x+1=\\=2+36x^6-36x^5-15x^4+24x^3-2x^2-4x+1$

$2+\left(3x^2-1\right)^2\left(2x-1\right)^2=\\=2+\left(9x^4-6x^2+1\right)\left(2x-1\right)^2=\\=2+\left(9x^4-6x^2+1\right)\left(4x^2-4x+1\right)=\\=2+9x^4\cdot \:4x^2+9x^4\left(-4x\right)+9x^4\cdot \:1+\left(-6x^2\right)\cdot \:4x^2+\left(-6x^2\right)\left(-4x\right)+\left(-6x^2\right)\cdot \:1+1\cdot \:4x^2+1\cdot \left(-4x\right)+1\cdot \:1=\\=2+9\cdot \:4x^4x^2-9\cdot \:4x^4x+9x^4-6\cdot \:4x^2x^2+6\cdot \:4x^2x-6x^2+4x^2-4x+1=\\=2+36x^6-36x^5-15x^4+24x^3-2x^2-4x+1=\\=36x^6-36x^5-15x^4+24x^3-2x^2-4x+3$

El resultado queda:

$f(x)=\frac{36x^6-36x^5-15x^4+24x^3-2x^2-4x+3}{(2x-1)^{2}}$