Cho $\frac{a}{b}$ = $\frac{c}{d}$ $\neq$ ± 1 và c $\neq$ 0. CMR: a)( $\frac{a-b}{c-d}$ ) ^2= $\frac{ab}{cd}$ b) (a+b)^2020/(c+d)^2020=a^2020-b^202

Cho $\frac{a}{b}$ = $\frac{c}{d}$ $\neq$ ± 1 và c $\neq$ 0. CMR:
a)( $\frac{a-b}{c-d}$ ) ^2= $\frac{ab}{cd}$
b) (a+b)^2020/(c+d)^2020=a^2020-b^2020/c^2020-d^2020

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  1. Đáp án:

    $\begin{array}{l}
    a)\dfrac{a}{b} = \dfrac{c}{d} = k \Rightarrow \left\{ \begin{array}{l}
    a = bk\\
    c = dk
    \end{array} \right.\\
    {\left( {\dfrac{{a – b}}{{c – d}}} \right)^2} = {\left( {\dfrac{{bk – b}}{{dk – d}}} \right)^2} = {\left( {\dfrac{b}{d}} \right)^2}\\
    \dfrac{{ab}}{{cd}} = \dfrac{{b.k.b}}{{d.k.d}} = \dfrac{{{b^2}}}{{{d^2}}} = {\left( {\dfrac{b}{d}} \right)^2}\\
     \Rightarrow {\left( {\dfrac{{a – b}}{{c – d}}} \right)^2} = \dfrac{{ab}}{{cd}}\\
    b)\dfrac{{{{\left( {a + b} \right)}^{2020}}}}{{{{\left( {c + d} \right)}^{2020}}}} = \dfrac{{{{\left( {bk + b} \right)}^{2020}}}}{{{{\left( {dk + d} \right)}^{2020}}}} = \dfrac{{{b^{2020}}}}{{{d^{2020}}}}\\
    \dfrac{{{a^{2020}} – {b^{2020}}}}{{{c^{2020}} – {d^{2020}}}} = \dfrac{{{b^{2020}}.{k^{2020}} – {b^{2020}}}}{{{d^{2020}}.{k^{2020}} – {d^{2020}}}} = \dfrac{{{b^{2020}}}}{{{d^{2020}}}}\\
     \Rightarrow \dfrac{{{{\left( {a + b} \right)}^{2020}}}}{{{{\left( {c + d} \right)}^{2020}}}} = \dfrac{{{a^{2020}} – {b^{2020}}}}{{{c^{2020}} – {d^{2020}}}}
    \end{array}$

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