Đáp án:
Ta có :
A = $\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1}{10}$ + ... + $\frac{1}{x(2x+1)}$ = $\frac{1}{10}$
=> A = $\frac{2}{6}$ + $\frac{2}{12}$ + $\frac{2}{20}$ + ... + $\frac{2}{2x(2x+1)}$ = $\frac{1}{10}$
=> A = 2.($\frac{1}{6}$ + $\frac{1}{12}$ + $\frac{1}{20}$ + ... + $\frac{1}{2x(2x+1) })$ = $\frac{1}{10}$
=> A = 2 . ( $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + $\frac{1}{4.5}$ + ... + $\frac{1}{x(2x+1)})$ = $\frac{1}{10}$
=> A = 2. ( $\frac{1}{2}$ - $\frac{1}{3}$ +$\frac{1}{3}$ - $\frac{1}{4}$ + $\frac{1}{4}$ -$\frac{1}{5}$ + ... +$\frac{1}{2x}$ - $\frac{1}{2x+1})$ = $\frac{1}{10}$
=> A = 2.($\frac{1}{2}$ - $\frac{1}{2x+1})$ ) = $\frac{1}{10}$
=> A = 2.$\frac{2x-1}{2(2x+1)}$ = $\frac{1}{10}$
=> A = $\frac{2x-1}{2x+1}$ = $\frac{1}{10}$
=> 2x - 1 = 1 => x = 1
=> $\frac{2x-1}{2x+1}$ = $\frac{1}{3}$ $\neq$ $\frac{1}{10}$
=> không có giá trị x thỏa mãn
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