Toán Lớp 8: $\frac{7}{8x^2-18}$ ${+}$ $\frac{1}{2x^2+3x}$ ${-}$ $\frac{1}{4x-6}$
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Toán Lớp 8: $\frac{7}{8x^2-18}$ ${+}$ $\frac{1}{2x^2+3x}$ ${-}$ $\frac{1}{4x-6}$
Comments ( 1 )
=$\frac{7}{2(2x-3)(2x+3)}$+$\frac{1}{x(2x+3)}$- $\frac{1}{2(2x-3)}$
=$\frac{7x}{2x(2x-3)(2x+3)}$+ $\frac{2(2x-3)}{2x(2x+3)(2x-3)}$-$\frac{x(2x+3)}{2x(2x-3)(2x+3)}$
=$\frac{7x+2(2x-3)-x(2x+3)}{2x(2x-3)(2x+3)}$
=$\frac{7x+4x-6-2x^{2}-3x}{2x(2x-3)(2x+3)}$
=$\frac{8x-6-2x^{2}}{2x(2x-3)(2x+3)}$
=$\frac{2(4x-3-x^{2})}{2x(2x-3)(2x+3)}$ = $\frac{4x-3-x^{2}}{x(2x-3)(2x+3)}$