by (1,5đ) Quy đồng mẫu thức các phân thức sau:
a) \(\frac{a}{{{{(a + b)}^2}}}\) và \(\frac{b}{{{{(a – b)}^2}}}\)
b) \(\frac{{x – 2}}{{x{y^2}}}\)và \(\frac{{1 – x}}{{12{x^3}{y^4}}}\)
c) \(\frac{{7x – 1}}{{2{x^2} + 6x}}\) và \(\frac{{5 – x}}{{{x^2} – 9}}\)
Lời giải chi tiết:
a) \(\frac{a}{{{{(a + b)}^2}}}\) và \(\frac{b}{{{{(a – b)}^2}}}\).
MTC: \({(a + b)^2}{(a – b)^2}\).
\(\begin{array}{l}\frac{a}{{{{(a + b)}^2}}} = \frac{{a{{(a – b)}^2}}}{{{{(a + b)}^2}{{(a – b)}^2}}}\\\frac{b}{{{{(a – b)}^2}}} = \frac{{b{{(a + b)}^2}}}{{{{(a + b)}^2}{{(a – b)}^2}}}\end{array}\)
b) \(\frac{{x – 2}}{{x{y^2}}}\) và \(\frac{{1 – x}}{{12{x^3}{y^4}}}\)
MTC: \(12{x^3}{y^4}\)
\(\begin{array}{l}\frac{{x – 2}}{{x{y^2}}} = \frac{{12{x^2}{y^2}(x – 2)}}{{12{x^3}{y^4}}} = \frac{{12{x^3}{y^2} – 24{x^2}{y^2}}}{{12{x^3}{y^4}}}\\\frac{{1 – x}}{{12{x^3}{y^4}}} = \frac{{1 – x}}{{12{x^3}{y^4}}}\end{array}\)
c) \(\frac{{7x – 1}}{{2{x^2} + 6x}}\) và \(\frac{{5 – x}}{{{x^2} – 9}}\)
Ta có:
\(2{x^2} + 6x = 2x(x + 3)\)
\({x^2} – 9 = (x + 3)(x – 3)\)
MTC: \(2x(x + 3)(x – 3)\)
\(\begin{array}{l}\frac{{7x – 1}}{{2{x^2} + 6x}} = \frac{{(7x – 1)(x – 3)}}{{2x(x – 3)(x + 3)}} = \frac{{7{x^2} – 22x + 3}}{{2x(x – 3)(x + 3)}}\\\frac{{5 – x}}{{{x^2} – 9}} = \frac{{2x(5 – x)}}{{2x(x – 3)(x + 3)}} = \frac{{ – 2{x^2} + 10x}}{{2x(x – 3)(x + 3)}}\end{array}\)
