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Đáp án: $x=\dfrac{24}{5}$
Giải thích các bước giải:
Bài 25: $2x+\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}=10$
$\to 2x+1+\dfrac{1}{6}+1+\dfrac{1}{12}+1+\dfrac{1}{20}+1+\dfrac{1}{30}+1+\dfrac{1}{42}+1+\dfrac{1}{56}+1+\dfrac{1}{72}+1+\dfrac{1}{90}=10$
$\to 2x+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}=2$
$\to 2x+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}=10$
$\to 2x+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=10$
$\to 2x+\dfrac{1}{2}-\dfrac{1}{10}=10$
$\to 2x=10-\dfrac{1}{2}+\dfrac{1}{10}$
$\to 2x=\dfrac{48}{5}$
$\to x=\dfrac{48}{5}:2$
$\to x=\dfrac{24}{5}$
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