Cho ab=cd.Chứng minh rằng : 7a2+3ab11a2-8b2=7c2+3cd11c2-8d2.Các anh chị giúp em với ạ
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\begin{array}{l} \frac{a}{b} = \frac{c}{d} = k\left( {b;d \ne 0} \right)\\ = > a = bk;c = dk\\ \frac{{7{a^2} + 3ab}}{{11{a^2} - 8{b^2}}} = \frac{{7{{\left( {bk} \right)}^2} + 3bk.b}}{{11{{\left( {bk} \right)}^2} - 8{b^2}}} = \frac{{{b^2}\left( {7{k^2} + 3k} \right)}}{{{b^2}\left( {11{k^2} - 8} \right)}} = \frac{{7{k^2} + 3k}}{{11{k^2} - 8}}\\ \frac{{7{c^2} + 3cd}}{{11{c^2} - 8{d^2}}} = \frac{{7{{\left( {dk} \right)}^2} + 3dk.d}}{{11{{\left( {dk} \right)}^2} - 8{d^2}}} = \frac{{{d^2}\left( {7{k^2} + 3k} \right)}}{{{d^2}\left( {11{k^2} - 8} \right)}} = \frac{{7{k^2} + 3k}}{{11{k^2} - 8}}\\ vay:\frac{{7{a^2} + 3ab}}{{11{a^2} - 8{b^2}}} = \frac{{7{c^2} + 3cd}}{{11{c^2} - 8{d^2}}} \end{array}
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