1. Khối 7
  2. Toán 7 - Cơ bản và nâng cao - Tập một
  3. Chương I SỐ HỮU TỈ
  4. §2. Cộng, trừ số hữu tỉ
  5. Mức độ nâng cao

  • Ví dụ  5∗~5^{*} 5∗. Tính: A=73.4−94.5+115.6−136.7+157.8−178.9+199.10A=\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}A=3.47​−4.59​+5.611​−6.713​+7.815​−8.917​+9.1019​.
    Bài giải
    A=6+112−10−120+10+130−14−142+14+156−18−172+18+190=612+112−1020+120+1030+130−1442+142+1456+156−1872+172+1890+190=12+112−12+120+13+130−13+142+14+156−14+172+15+190=112+120+130+142+156+172+190+15=13−14+14−15+15−16+16−17+17−18+18−19+19−110+15=13−110+15=1030−330+630=1330. \begin{aligned} A &=\dfrac{6+1}{12}-\dfrac{10-1}{20}+\dfrac{10+1}{30}-\dfrac{14-1}{42}+\dfrac{14+1}{56}-\dfrac{18-1}{72}+\dfrac{18+1}{90} \\ &=\dfrac{6}{12}+\dfrac{1}{12}-\dfrac{10}{20}+\dfrac{1}{20}+\dfrac{10}{30}+\dfrac{1}{30}-\dfrac{14}{42}+\dfrac{1}{42}+\dfrac{14}{56}+\dfrac{1}{56}-\dfrac{18}{72}+\dfrac{1}{72}+\dfrac{18}{90}+\dfrac{1}{90} \\ &=\dfrac{1}{2}+\dfrac{1}{12}-\dfrac{1}{2}+\dfrac{1}{20}+\dfrac{1}{3}+\dfrac{1}{30}-\dfrac{1}{3}+\dfrac{1}{42}+\dfrac{1}{4}+\dfrac{1}{56}-\dfrac{1}{4}+\dfrac{1}{72}+\dfrac{1}{5}+\dfrac{1}{90} \\ &=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{5} \\ &=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{5} \\ &=\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{5}=\dfrac{10}{30}-\dfrac{3}{30}+\dfrac{6}{30}=\dfrac{13}{30} . \end{aligned} A​=126+1​−2010−1​+3010+1​−4214−1​+5614+1​−7218−1​+9018+1​=126​+121​−2010​+201​+3010​+301​−4214​+421​+5614​+561​−7218​+721​+9018​+901​=21​+121​−21​+201​+31​+301​−31​+421​+41​+561​−41​+721​+51​+901​=121​+201​+301​+421​+561​+721​+901​+51​=31​−41​+41​−51​+51​−61​+61​−71​+71​−81​+81​−91​+91​−101​+51​=31​−101​+51​=3010​−303​+306​=3013​.​

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