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(1) Prove that\operatorname{se}\left(\frac{51}{80}\right)+\log \left(\frac{44}{85}\right)-\log \left(\frac{99}{160}\right)-\log \left(\frac{8}{15}\right)=0 |
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Answer» 51=17×380=16×544=11×2×285=17×599=11×3×3160=16×5×28=2×2×215=3×5log(a/b)=loga-logblogab=loga+logblog17+log3-log16-log5+log11+log2+log2-log17-log5-log11-log3-log3+log16+log5+log2-log2-log2-log2+log3+log5=0 |
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