Question

a) 75% - \(\dfrac{3}{4}\) : \(x\)\(^2\) = \(-2\dfrac{1}{4}\)

b) 8\(x^4\) + \(x\) = 0

Answer 1

a) \(75\%-\dfrac{3}{4}:x^2=-2\dfrac{1}{4}\)

\(\Rightarrow\dfrac{3}{4}-\dfrac{3}{4}:x^2=-\dfrac{9}{4}\)

\(\Rightarrow\dfrac{3}{4}:x^2=\dfrac{3}{4}+\dfrac{9}{4}=3\)

\(\Rightarrow x^2=\dfrac{3}{4}:3=\dfrac{3}{4}.\dfrac{1}{3}=\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^2\)

\(\Rightarrow x=-\dfrac{1}{2};x=\dfrac{1}{2}\) hay \(x\in\left\{-\dfrac{1}{2};\dfrac{1}{2}\right\}\)

b) \(8x^4+x=0\)

\(\Rightarrow x\left(8x^3+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\8x^3+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^3=-\dfrac{1}{8}=\left(-\dfrac{1}{2}\right)^3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{0;-\dfrac{1}{2}\right\}\)