Tìm x biết |x+4/5|-1/2=0

1)

a) \(|x-3,5|=7,5\)

\(\Rightarrow x-3,5=7,5\)

hay \(x-3,5=-7,5\)

TH1 : \(x-3,5=7,5\Rightarrow x=7,5+3,5=11\)

TH2 : \(x-3,5=-7,5\Rightarrow x=-7,5+3,5=-4\)

b) \(|x+\dfrac{4}{5}|-\dfrac{1}{2}=0\)

\(\Rightarrow\left(x+\dfrac{4}{5}\right)-\dfrac{1}{2}=0\) (chỉ có 1 TH vì số 0 ko phải dương or âm)

\(\left(x+\dfrac{4}{5}\right)=0+\dfrac{1}{2}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}-\dfrac{4}{5}=\dfrac{5-8}{10}=\dfrac{-3}{10}\)

c) \(3,6-|x-0,4|=0\)

\(\Rightarrow3,6-\left(x-0,4\right)=0\) ( giải thích giống câu b )

\(\Rightarrow-\left(x-0,4\right)=0-3,6\)

\(\Rightarrow-\left(x-0,4\right)=-3,6\)

\(\Rightarrow-x+0,4=-3,6\) ( Phá dấu )

\(\Rightarrow-x=-3,6-0,4=-3,6+\left(-0,4\right)=-4\)

\(\Rightarrow x=4\)

d) \(-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{-5}{9}\)

\(\Rightarrow-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{-5}{9}\)

hay \(\Rightarrow-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{5}{9}\)

TH1 : \(-\dfrac{5}{12}:\left(-\dfrac{5}{6}:x\right)=\dfrac{-5}{9}\Rightarrow\left(-\dfrac{5}{6}:x\right)=-\dfrac{5}{12}:\left(-\dfrac{5}{9}\right)\)

\(\Rightarrow\left(-\dfrac{5}{6}:x\right)=\dfrac{5}{12}.\dfrac{9}{5}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(\Rightarrow x=-\dfrac{5}{6}:\dfrac{3}{4}=-\dfrac{5.4}{6.3}=-\dfrac{5.2}{3.3}=-\dfrac{10}{9}\)

TH2 : \(\Rightarrow-\dfrac{5}{12}:\left(-\dfrac{5}{6}:x\right)=\dfrac{5}{9}\)

\(\Rightarrow\)\(\left(-\dfrac{5}{6}:x\right)=-\dfrac{5}{12}:\dfrac{5}{9}=-\dfrac{5.9}{12.5}=-\dfrac{9}{12}=-\dfrac{3}{4}\)

\(\Rightarrow x=-\dfrac{5}{6}:\left(-\dfrac{3}{4}\right)=\dfrac{5}{6}.\dfrac{4}{3}=\dfrac{10}{9}\)

Vậy x = ....

e)

Vì \(|x-3,5|\ge0;|4,5-x|\ge0\) với mọi x

Do đó : \(|x-3,5|+|4,5-x|=0\)

\(\Rightarrow|x-3,5|=0;|4,5-x|=0\)

\(\Rightarrow x-3,5=0\) và \(4,5-x=0\)

\(\Rightarrow x=0+3,5=3,5\) và \(-x=0+4,5=4,5\Rightarrow x=-4,5\)

( không đồng thời xảy ra)

\(\Rightarrow\) Không tồn tại x thuộc Q để \(|x-3,5|+|4,5-x|=0\)