Tìm x ∈ N , x ≥ 2 , biết : 1/2.4+1/4.6+…+1/(2x-2).2x=1/8 15/10/2021 Bởi Emery Tìm x ∈ N , x ≥ 2 , biết : 1/2.4+1/4.6+…+1/(2x-2).2x=1/8
Đáp án: $x=2$ Giải thích các bước giải: $\dfrac{1}{2.4}+\dfrac{1}{4.6}+…+\dfrac{1}{(2x-2).2x}=\dfrac{1}{8}\\\Leftrightarrow 2.\left ( \dfrac{1}{2.4}+\dfrac{1}{4.6}+…+\dfrac{1}{(2x-2).2x} \right )=\dfrac{1}{8}.2\\\Leftrightarrow \dfrac{2}{2.4}+\dfrac{2}{4.6}+…+\dfrac{2}{(2x-2).2x}=\dfrac{1}{4}\\\Leftrightarrow \dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+…+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\\\Leftrightarrow \dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\\\Leftrightarrow 1-\dfrac{1}{x}=\dfrac{1}{2}\\\Leftrightarrow \dfrac{1}{x}=1-\dfrac{1}{2}=\dfrac{1}{2}\\\Leftrightarrow x=2$ Bình luận
Đáp án:
$x=2$
Giải thích các bước giải:
$\dfrac{1}{2.4}+\dfrac{1}{4.6}+…+\dfrac{1}{(2x-2).2x}=\dfrac{1}{8}\\
\Leftrightarrow 2.\left ( \dfrac{1}{2.4}+\dfrac{1}{4.6}+…+\dfrac{1}{(2x-2).2x} \right )=\dfrac{1}{8}.2\\
\Leftrightarrow \dfrac{2}{2.4}+\dfrac{2}{4.6}+…+\dfrac{2}{(2x-2).2x}=\dfrac{1}{4}\\
\Leftrightarrow \dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+…+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\\
\Leftrightarrow \dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\\
\Leftrightarrow 1-\dfrac{1}{x}=\dfrac{1}{2}\\
\Leftrightarrow \dfrac{1}{x}=1-\dfrac{1}{2}=\dfrac{1}{2}\\
\Leftrightarrow x=2$