Question

доведіть що значення виразу не залежить від значення змінної x

Answer 1

Объяснение:

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