Question

(4 phần 3 cộng x) mũ 2 = 1 phần 2

Answer 1

x = (3-4\(\sqrt{2}\))/3\(\sqrt{2}\)

Answer 2

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

\(\sqrt{\left(\dfrac{4}{3}+x\right)^2}=\sqrt{\dfrac{1}{2}}\)

\(\left|\dfrac{4}{3}+x\right|=\dfrac{\sqrt{2}}{2}\)

\(\dfrac{4}{3}+x=\pm\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{4}{3}+x=\dfrac{\sqrt{2}}{2}\\\dfrac{4}{3}-x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{\sqrt{2}}{2}-\dfrac{4}{3}=\dfrac{-8+3\sqrt{2}}{6}\\x=-\dfrac{\sqrt{2}}{2}-\dfrac{4}{3}=-\dfrac{8+3\sqrt{2}}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{-8+3\sqrt{2}}{6};-\dfrac{8+3\sqrt{2}}{6}\right\}\).

Answer 3

\(\left(\dfrac{4}{3}+x\right)^2=\dfrac{1}{2}\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{4}{3}+x=\dfrac{\sqrt{2}}{2}\\\dfrac{4}{3}+x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{2}}{2}-\dfrac{4}{3}\\x=\dfrac{-\sqrt{2}}{2}-\dfrac{4}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt{2}-8}{6}\\x=-\dfrac{3\sqrt{2}+8}{6}\end{matrix}\right.\)