a) (x-1/3)(2x+4)=0 b) (x-2/5)^2=9/25 c) (x-1/2)(2x+3)<0 d) 5/x-1<0 e) x+2/x-3>1

0 bình luận về “a) (x-1/3)(2x+4)=0 b) (x-2/5)^2=9/25 c) (x-1/2)(2x+3)<0 d) 5/x-1<0 e) x+2/x-3>1”

1. a,(x-$\frac{1}{3}$)(2x+4)=0

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

   b,Xét x-2/5=3/5

   x=1

   Xét x-2/5=-3/5

   x=-1/5

   c,Có 2x+3>x-1/2

   ⇒x-1/2<0

   ⇒x<1/2

   d,5/x-1<0

   ⇔5/x<1

   ⇔5<x

   e,x+2/x-3>1

   ⇔x-3+5/x-3>1

   ⇔1+5/x-3>1

   ⇔5/x-3>0

   ⇔x-3>0

   ⇔x>3

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2. $a$) `(x-1/3)(2x+4)=0`

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

     Vậy $x$ $∈$ `{1/3;-2}`

   $b$) `(x-2/5)^2=9/25`

   `⇔ (x-2/5)^2 = (± 3/5)^2`

   `⇔ x – 2/5 = ± 3/5`

   `⇒` \(\left[ \begin{array}{l}x=1\\x=-\dfrac{1}{5}\end{array} \right.\) 

    Vậy $x$ $∈$ `{1;-1/5}`

   $c$) `(x-1/2)(2x+3)<0`

   `⇒ x-1/2;2x+3` khác dấu

   $TH1$. $\left\{\begin{matrix}x-1 > 0 & \\ 2x+ 3 < 0& \end{matrix}\right.$ $⇒$ $KTM$

   $TH2$. $\left\{\begin{matrix}x-1 < 0 & \\ 2x+ 3 > 0& \end{matrix}\right.$ $⇒$ `-3/2 < x < 1`

    Vậy `-3/2 < x < 1`

   $d$) `5/{x-1} < 0`

   `⇒ x-1 < 0`

   `⇒ x < 1`

     Vậy `x<1`

   $e$) `{x+2}/{x-3} > 1`

   `⇔ {x-3+5}/{x-3} > 1`

   `⇔ 1 + 5/{x-3} > 1`

   `⇒ 5/{x-3} > 0`

   `⇔ x -3 > 0`

   `⇔ x > 3`

     Vậy `x>3`.

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