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

Question

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

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Ariana 1 tháng 2021-08-07T20:21:52+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-08-07T20:23:18+00:00

    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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    0
    2021-08-07T20:23:45+00:00

    $a$) `(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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