化简为最简的单分式形式。 $\dfrac{3t^2}{w} \cdot \dfrac{9t^2}{w^3}$ 答案 ------------------------------------------------ $\dfrac{3}{y-1} - \dfrac{5}{y+6}$ 答案 ... 化简 $\displaystyle \dfrac{2v^2 - 5v - 12}{16 - v^2}$ 答案 ...

1. Multiply the algebraic fractions and combine like bases.

$\frac{3t^2}{w}\cdot\frac{9t^2}{w^3}=\frac{3\times 9\times t^2\times t^2}{w\times w^3}=\frac{27t^4}{w^4}$

1. Use the common denominator for the two algebraic fractions.

Common denominator: $(y-1)(y+6)$

$\frac{3}{y-1}-\frac{5}{y+6}=\frac{3(y+6)}{(y-1)(y+6)}-\frac{5(y-1)}{(y-1)(y+6)}$

2. Subtract the numerators and simplify.

$\frac{3(y+6)-5(y-1)}{(y-1)(y+6)}$

$=\frac{3y+18-5y+5}{(y-1)(y+6)}$

$=\frac{13-2y}{(y-1)(y+6)}$

1. Factorise the quadratic numerator.

$2v^2-5v-12=(2v+3)(v-4)$

2. Factorise the denominator using difference of squares and adjust the sign to match a common factor.

$16-v^2=(4-v)(4+v)=-(v-4)(v+4)$

3. Cancel the common factor and simplify the sign.

$\frac{2v^2-5v-12}{16-v^2}=\frac{(2v+3)(v-4)}{-(v-4)(v+4)}=-\frac{2v+3}{v+4}$