a) ( 1/3x – 8/3 ) . (2,5 + -7/5 ) = 0 b) ( -5/4x + 3,25 ) . (3/5 – (-5/2x)) = 0 c) ( x – 2/7 ) . ( x + 3/4 ) = 0 d) (2x + 1/5 ) . (-3/5x + 4/7 ) = 0

0 bình luận về “a) ( 1/3x – 8/3 ) . (2,5 + -7/5 ) = 0 b) ( -5/4x + 3,25 ) . (3/5 – (-5/2x)) = 0 c) ( x – 2/7 ) . ( x + 3/4 ) = 0 d) (2x + 1/5 ) . (-3/5x + 4/7 ) = 0”

1. Đáp án:

   Đề và xem lại câu `a,` ở đây nka.

   `a,`

   `(1/3x – 8/3) (2,5 + (-7)/5 ÷ x) = 0`

   `->` \(\left[ \begin{array}{l}\dfrac{1}{3}x-\dfrac{8}{3}=0\\2,5 + \dfrac{-7}{5}÷x=0\end{array} \right.\) 

   `->` \(\left[ \begin{array}{l}\dfrac{1}{3}x=\dfrac{8}{3}\\ \dfrac{-7}{5}÷x=-2,5\end{array} \right.\) 

   `->` \(\left[ \begin{array}{l}x=8\\x=\dfrac{14}{25}\end{array} \right.\) 

   Vậy `x ∈ {8;14/25}`

    

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2. `a,` `(1/3x-8/3).(2,5x+{-7}/5)=0`

   `⇒(1/3x-8/3).(5/2x-7/5)=0`

   $⇒\left[ \begin{array}{l}\dfrac13x-\dfrac{8}{3}=0\\\dfrac52x+\dfrac{-7}5=0\end{array} \right.$

   $⇒\left[ \begin{array}{l}\dfrac13x=\dfrac{8}{3}\\\dfrac52x=\dfrac75\end{array} \right.$

   $⇒\left[ \begin{array}{l}x=8\\x=\dfrac{14}{25}\end{array} \right.$

   `b,` `({-5}/4x+3,25).(3/5-{-5}/2x)=0`

   `⇒({-5}/4x+13/4).(3/5-{-5}/2x)=0`

   $⇒\left[ \begin{array}{l}\dfrac{-5}4x+\dfrac{13}{4}=0\\\dfrac35-\dfrac{-5}2x=0\end{array} \right.$

   $⇒\left[ \begin{array}{l}\dfrac{-5}4x=\dfrac{-13}{4}\\\dfrac35=\dfrac{-5}2x\end{array} \right.$

   $⇒\left[ \begin{array}{l}x=\dfrac{13}5\\x=\dfrac{-6}{25}\end{array} \right.$

   `c,` `( x-2/7).(x+3/4)=0`

   $⇒\left[ \begin{array}{l}x-\dfrac27=0\\x+\dfrac34=\end{array} \right.$

   $⇒\left[ \begin{array}{l}x=\dfrac27\\x=\dfrac{-3}4\end{array} \right.$

   `d,` `(2x+1/5).(-3/5x+4/7)=0`

   $⇒\left[ \begin{array}{l}2x+\dfrac15=0\\\dfrac{-3}5x+\dfrac47=0\end{array} \right.$

   $⇒\left[ \begin{array}{l}2x=\dfrac{-1}5\\\dfrac{-3}5=\dfrac{-4}7\end{array} \right.$

   $⇒\left[ \begin{array}{l}x=\dfrac{-1}{10}\\x=\dfrac{20}{21}\end{array} \right.$

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