(1,5đ) Quy đồng mẫu thức các phân...

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

a) \(\dfrac{a}{{{{(a + b)}^2}}}\) và \(\dfrac{b}{{{{(a - b)}^2}}}\).

MTC: \({(a + b)^2}{(a - b)^2}\).

\(\begin{array}{l}\dfrac{a}{{{{(a + b)}^2}}} = \dfrac{{a{{(a - b)}^2}}}{{{{(a + b)}^2}{{(a - b)}^2}}}\\\dfrac{b}{{{{(a - b)}^2}}} = \dfrac{{b{{(a + b)}^2}}}{{{{(a + b)}^2}{{(a - b)}^2}}}\end{array}\)

b) \(\dfrac{{x - 2}}{{x{y^2}}}\) và \(\dfrac{{1 - x}}{{12{x^3}{y^4}}}\)

MTC: \(12{x^3}{y^4}\)

\(\begin{array}{l}\dfrac{{x - 2}}{{x{y^2}}} = \dfrac{{12{x^2}{y^2}(x - 2)}}{{12{x^3}{y^4}}} = \dfrac{{12{x^3}{y^2} - 24{x^2}{y^2}}}{{12{x^3}{y^4}}}\\\dfrac{{1 - x}}{{12{x^3}{y^4}}} = \dfrac{{1 - x}}{{12{x^3}{y^4}}}\end{array}\)

 c) \(\dfrac{{7x - 1}}{{2{x^2} + 6x}}\) và \(\dfrac{{5 - x}}{{{x^2} - 9}}\)

Ta có:

\(2{x^2} + 6x = 2x(x + 3)\)

\({x^2} - 9 = (x + 3)(x - 3)\)

MTC: \(2x(x + 3)(x - 3)\)

\(\begin{array}{l}\dfrac{{7x - 1}}{{2{x^2} + 6x}} = \dfrac{{(7x - 1)(x - 3)}}{{2x(x - 3)(x + 3)}} = \dfrac{{7{x^2} - 22x + 3}}{{2x(x - 3)(x + 3)}}\\\dfrac{{5 - x}}{{{x^2} - 9}} = \dfrac{{2x(5 - x)}}{{2x(x - 3)(x + 3)}} = \dfrac{{ - 2{x^2} + 10x}}{{2x(x - 3)(x + 3)}}\end{array}\)