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

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.