by (1,5đ) Rút gọn các phân thức sau:
a) \(A = \frac{{3\left| {x – 2} \right| – 5\left| {x – 6} \right|}}{{4{x^2} – 36{\text{x}} + 81}}\) với \(2 < x < 6\).
b) \(B = \frac{{x|x – 2|}}{{{x^3} – 5{x^2} + 6x}}\)
Lời giải chi tiết:
a) \(A = \frac{{3\left| {x – 2} \right| – 5\left| {x – 6} \right|}}{{4{x^2} – 36{\text{x}} + 81}}\) với \(2 < x < 6\).
Với \(2 < x 0\) và \(x – 6 < 0.\)
\( \Rightarrow |x – 2| = x – 2\) và \(|x – 6| = 6 – x.\)
\(A = \frac{{3\left| {x – 2} \right| – 5\left| {x – 6} \right|}}{{4{x^2} – 36{\text{x}} + 81}} = \frac{{3(x – 2) – 5(6 – x)}}{{{{(2x – 9)}^2}}} = \frac{{3x – 6 – 30 + 5x}}{{{{(2x – 9)}^2}}} = \frac{{8x – 36}}{{{{(2x – 9)}^2}}} = \frac{{4(2x – 9)}}{{{{(2x – 9)}^2}}} = \frac{4}{{2x – 9}}.\)
b) \(B = \frac{{x|x – 2|}}{{{x^3} – 5{x^2} + 6x}} = \frac{{x|x – 2|}}{{x({x^2} – 5x + 6)}} = \frac{{x|x – 2|}}{{x({x^2} – 2x – 3x + 6)}} = \frac{{x|x – 2|}}{{x{\text{[}}x(x – 2) – 3(x – 2){\text{]}}}} = \frac{{x|x – 2|}}{{x(x – 2)(x – 3)}}\) .
Nếu \(x – 2 \geqslant 0 \Leftrightarrow x \geqslant 2\) thì \(|x – 2| = x – 2 \Rightarrow B = \frac{{x(x – 2)}}{{x(x – 2)(x – 3)}} = \frac{1}{{x – 3}}.\)
Nếu \(x – 2 < 0 \Leftrightarrow x < 2\) thì \(|x – 2| = 2 – x \Rightarrow B = \frac{{x(2 – x)}}{{x(x – 2)(x – 3)}} = \frac{{x(x – 2)}}{{x(x – 2)(3 – x)}} = \frac{1}{{3 – x}}.\)
