(2đ) Quy đồng mẫu thức các thức sau: a) \(A = \frac{5}{{2x – 4}};B = \frac{4}{{3x – 9}};C = \frac{7}{{10 – 5x}}\) b) \(D = \frac{{{x^2}}}{{{x^2} – 1}};E = \frac{{3x – 1}}{{{x^3} + 2{x^2} + x}};F = \frac{{2x + 1}}{{{x^3}}}\)

(2đ) Quy đồng mẫu thức các thức sau:

a) \(A = \frac{5}{{2x – 4}};B = \frac{4}{{3x – 9}};C = \frac{7}{{10 – 5x}}\)

b) \(D = \frac{{{x^2}}}{{{x^2} – 1}};E = \frac{{3x – 1}}{{{x^3} + 2{x^2} + x}};F = \frac{{2x + 1}}{{{x^3}}}\)

Lời giải chi tiết:

\(\begin{array}{l}

a)A = \frac{5}{{2x – 4}};B = \frac{4}{{3x – 9}};C = \frac{7}{{10 – 5x}}\\

2x – 4 = 2(x – 2)\\

3x – 9 = 3(x – 3)\\

10 – 5x = 5(2 – x)\\

MTC:2.3.5(x – 2)(x – 3) = 30(x – 2)(x – 3)\\

A = \frac{5}{{2x – 4}} = \frac{{5.3.5(x – 3)}}{{30(x – 2)(x – 3)}} = \frac{{75(x – 3)}}{{30(x – 2)(x – 3)}} = \frac{{75x – 225}}{{30(x – 2)(x – 3)}}\\

B = \frac{4}{{3x – 9}} = \frac{{4.2.5(x – 2)}}{{30(x – 2)(x – 3)}} = \frac{{40(x – 2)}}{{30(x – 2)(x – 3)}} = \frac{{40x – 80}}{{30(x – 2)(x – 3)}}\\

C = \frac{7}{{10 – 5x}} = \frac{{ – 7}}{{5(x – 2)}} = \frac{{ – 7.2.3(x – 3)}}{{30(x – 2)(x – 3)}} = \frac{{ – 42(x – 3)}}{{30(x – 2)(x – 3)}} = \frac{{ – 42x + 126}}{{30(x – 2)(x – 3)}}\\

b)D = \frac{{{x^2}}}{{{x^2} – 1}};E = \frac{{3x – 1}}{{{x^3} + 2{x^2} + x}};F = \frac{{2x + 1}}{{{x^3}}}\\

{x^2} – 1 = (x – 1)(x + 1)\\

{x^3} + 2{x^2} + x = x({x^2} + 2x + 1) = x{(x + 1)^2}\\

MTC:{x^3}{(x + 1)^2}(x – 1)\\

D = \frac{{{x^2}}}{{{x^2} – 1}} = \frac{{{x^2}}}{{(x – 1)(x + 1)}} = \frac{{{x^2}.{x^3}\left( {x + 1} \right)}}{{{x^3}{{(x + 1)}^2}(x – 1)}} = \frac{{{x^5}(x + 1)}}{{{x^3}{{(x + 1)}^2}(x – 1)}}\\

E = \frac{{3x – 1}}{{{x^3} + 2{x^2} + x}} = \frac{{3x – 1}}{{x{{(x + 1)}^2}}} = \frac{{(3x – 1){x^2}(x – 1)}}{{{x^3}{{(x + 1)}^2}(x – 1)}}\\

F = \frac{{2x + 1}}{{{x^3}}} = \frac{{(2x + 1)(x – 1){{(x + 1)}^2}}}{{{x^3}{{(x + 1)}^2}(x – 1)}}

\end{array}\)