3. ( x − 2 ) = 18 − ( 6 + x ) 2 /5. ( x − 1/ 2 .x ) = 1 /5 + 2. x 2 /5 . ( 1 − 1/ 2 .x ) = 2 /5 . ( 1/ 3 + 10/ 3. x )

0 bình luận về “3. ( x − 2 ) = 18 − ( 6 + x ) 2 /5. ( x − 1/ 2 .x ) = 1 /5 + 2. x 2 /5 . ( 1 − 1/ 2 .x ) = 2 /5 . ( 1/ 3 + 10/ 3. x )”

1. Đáp án:

    a) \(x = \dfrac{9}{2}\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   a)3\left( {x – 2} \right) = 18 – 6 – x\\
    \to 3x – 6 = 12 – x\\
    \to 4x = 18\\
    \to x = \dfrac{9}{2}\\
   b)DK:x \ne 0\\
   \dfrac{2}{5}.\left( {\dfrac{{x – 1}}{{2x}}} \right) = \dfrac{1}{5} + 2x\\
    \to \dfrac{{2x – 2}}{{10x}} = \dfrac{{1 + 10x}}{5}\\
    \to 10x – 10 = 10x + 100{x^2}\\
    \to 100{x^2} =  – 10\left( l \right)\\
    \to x \in \emptyset \\
   c)DK:x \ne 0\\
   \dfrac{2}{5}.\left( {1 – \dfrac{1}{{2x}}} \right) = \dfrac{2}{5}.\left( {\dfrac{1}{3} + \dfrac{{10}}{{3x}}} \right)\\
    \to 1 – \dfrac{1}{{2x}} = \dfrac{1}{3} + \dfrac{{10}}{{3x}}\\
    \to \dfrac{{2x – 1}}{{2x}} = \dfrac{{x + 10}}{{3x}}\\
    \to \dfrac{{6x – 3}}{{6x}} = \dfrac{{2x + 20}}{{6x}}\\
    \to 6x – 3 = 2x + 20\\
    \to 4x = 23\\
    \to x = \dfrac{{23}}{4}
   \end{array}\)

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