Tìm x a, (2x – 3)² = 16 b, ( -1/3)³ . x = 1/81

Tìm x
a, (2x – 3)² = 16
b, ( -1/3)³ . x = 1/81

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  1. CHÚC BẠN HỌC TỐT!!!

    Trả lời:

    $a, (2x-3)^2=16$

    \(⇔\left[ \begin{array}{l}2x-3=4\\2x-3=-4\end{array} \right.\)\(⇔\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{array} \right.\)

     Vậy $S=\{\dfrac{7}{2};-\dfrac{1}{2}\}$

    $b, \bigg{(}-\dfrac{1}{3}\bigg{)}^3.x=\dfrac{1}{81}$

    $⇔-\dfrac{1}{27}.x=\dfrac{1}{81}$

    $⇔x=-\dfrac{1}{3}$

    Vậy $S=\{-\dfrac{1}{3}\}$.

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  2. a, `(2x – 3)² = 16`

    `=> 2x – 3 = 4 ` hoặc `2x – 3 = – 4`

    `=> 2x = 7` hoặc`2x = – 1`

    `=> x = 7/2` hoặc `x=-1/2`

    b, ` ( -1/3)³ . x = 1/81`

    `=> ( -1/3)³ . x = (1/3)^4`

    `=> x = -1/3`

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