Chứng minh rằng $\frac{a}{b}$ = $\frac{c}{d}$ thì: a) $\frac{a-b}{a+b}$ = $\frac{c-d}{c+d}$ b) $\frac{2a+5b}{3a-4b}$ = $\frac{2c+5d}{3c-4d}$ c) $\f

Đáp án:

$\begin{array}{l}
\dfrac{a}{b} = \dfrac{c}{d} = k \Rightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
a)\dfrac{{a – b}}{{a + b}} = \dfrac{{b.k – b}}{{b.k + b}} = \dfrac{{k – 1}}{{k + 1}}\\
\dfrac{{c – d}}{{c + d}} = \dfrac{{d.k – d}}{{d.k + d}} = \dfrac{{k – 1}}{{k + 1}}\\
 \Rightarrow \dfrac{{a – b}}{{a + b}} = \dfrac{{c – d}}{{c + d}}\\
b)\dfrac{{2a + 5b}}{{3a – 4b}} = \dfrac{{2.b.k + 5b}}{{3.b.k – 4b}} = \dfrac{{2k + 5}}{{3k – 4}}\\
\dfrac{{2c + 5b}}{{3c – 4d}} = \dfrac{{2.d.k + 5b}}{{3.d.k – 4d}} = \dfrac{{2k + 5}}{{3k – 4}}\\
 \Rightarrow \dfrac{{2a + 5b}}{{3a – 4b}} = \dfrac{{2c + 5b}}{{3c – 4d}}\\
c)\dfrac{{5a + 3b}}{{5a – 3b}} = \dfrac{{5.b.k + 3b}}{{5.b.k – 3b}} = \dfrac{{5k + 3}}{{5k – 3}}\\
\dfrac{{5c + 3d}}{{5c – 3d}} = \dfrac{{5.d.k + 3d}}{{5.d.k – 3d}} = \dfrac{{5k + 3}}{{5k – 3}}\\
 \Rightarrow \dfrac{{5a + 3b}}{{5a – 3b}} = \dfrac{{5c + 3d}}{{5c – 3d}}\\
d)\dfrac{{ab}}{{cd}} = \dfrac{{bk.b}}{{d.k.d}} = \dfrac{{{b^2}}}{{{d^2}}}\\
\dfrac{{{{\left( {a – b} \right)}^2}}}{{{{\left( {c – d} \right)}^2}}} = \dfrac{{{{\left( {bk – b} \right)}^2}}}{{{{\left( {dk – d} \right)}^2}}} = \dfrac{{{b^2}{{\left( {k – 1} \right)}^2}}}{{{d^2}{{\left( {k – 1} \right)}^2}}} = \dfrac{{{b^2}}}{{{d^2}}}\\
 \Rightarrow \dfrac{{ab}}{{cd}} = \dfrac{{{{\left( {a – b} \right)}^2}}}{{{{\left( {c – d} \right)}^2}}}
\end{array}$