Question

( \(\dfrac{1}{2}\): x + \(\dfrac{-3}{4}\)  )² -  \(\dfrac{25}{4}\)  = 0

Answer 1

=>(1/2:x-3/4-5/2)(1/2:x-3/4+5/2)=0

=>(1/2:x-13/4)*(1/2:x+7/4)=0

=>x=1/2:13/4=1/2*4/13=2/13 hoặc x=-1/2:7/4=-1/2*4/7=-4/14=-2/7

Answer 2

\(\Leftrightarrow\left(\dfrac{1}{2x}-\dfrac{3}{4}\right)^2=\dfrac{25}{4}=\left(\dfrac{5}{2}\right)^2\\ \Leftrightarrow\left(\dfrac{1}{2x}-\dfrac{3}{4}\right)=\dfrac{5}{2}\\ \Leftrightarrow\dfrac{1}{2x}=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{10}{4}+\dfrac{3}{4}=\dfrac{13}{4}\\ \Leftrightarrow2x=1:\dfrac{13}{4}=\dfrac{4}{13}\\ \Rightarrow x=\dfrac{4}{13}:2=\dfrac{4}{13.2}=\dfrac{4}{26}=\dfrac{2}{13}\)

Answer 3

`@` `\text {Ans}`

`\downarrow`

\(\left(\dfrac{1}{2}\div x+\dfrac{-3}{4}\right)^2-\dfrac{25}{4}=0\)

`=>`\(\left(\dfrac{1}{2}\div x+\dfrac{-3}{4}\right)^2=\dfrac{25}{4}\)

`=>`\(\left(\dfrac{1}{2}\div x+\dfrac{-3}{4}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)

`=>`\(\left[{}\begin{matrix}\dfrac{1}{2}\div x+\dfrac{-3}{4}=\dfrac{5}{2}\\\dfrac{1}{2}\div x+\dfrac{-3}{4}=-\dfrac{5}{2}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}\dfrac{1}{2}\div x=\dfrac{13}{4}\\\dfrac{1}{2}\div x=-\dfrac{7}{4}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{2}{13}\\x=-\dfrac{2}{7}\end{matrix}\right.\)

Vậy, `x \in {2/13; -2/7}.`

`\text {KaizuulvG}`