b: \(x^2-\dfrac{16}{25}=0\)
\(\Leftrightarrow x^2=\dfrac{16}{25}\)
hay \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)
b) \(x^2-\dfrac{16}{25}=0\Rightarrow x^2=\dfrac{16}{25}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
c) \(\dfrac{2}{5}-\left|\dfrac{1}{2}-x\right|=6\)
\(\Rightarrow\left|\dfrac{1}{2}-x\right|=-\dfrac{28}{5}\)(vô lý do \(\left|\dfrac{1}{2}-x\right|\ge0\))
Vậy \(S=\varnothing\)
b,x2=0+16/25
TH1:x=4/5
TH2:x=-4/5
c,|1/2-x|=-28/5(vì 2/5-6)
bỏ(vì vô lí)
