\(4,\dfrac{11x}{2x-3}-\dfrac{18-x}{2x-3}=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\\ 5,\dfrac{2x^3-x}{x-1}+\dfrac{x^3+3}{1-x}+\dfrac{2+x}{x-1}=\dfrac{2x^3-x+2+x}{x-1}-\dfrac{x^3+3}{x-1}=\dfrac{2x^3-x+2+x-x^3-3}{x-1}=\dfrac{x^3-1}{x-1}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}=x^2+x+1\)
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