=> 1/3x-1/4+x^2-9/16=0
=(1/3x+x/2)+....
các bước sau tự giải
Ta có: \(\hept{\begin{cases}\left(\frac{1}{3}x-\frac{1}{4}\right)^2\ge0;\forall x\\\left(x^2-\frac{9}{16}\right)^4\ge0;\forall x\end{cases}}\)\(\Rightarrow\left(\frac{1}{3}x-\frac{1}{4}\right)^2+\left(x^2-\frac{9}{16}\right)^4\ge0;\forall x\)
Do đó \(\left(\frac{1}{3}x-\frac{1}{4}\right)^2+\left(x^2-\frac{9}{16}\right)^4=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(\frac{1}{3}x-\frac{1}{4}\right)^2=0\\\left(x^2-\frac{9}{16}\right)^4=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\pm\frac{3}{4}\end{cases}}\)
\(\Leftrightarrow x=\frac{3}{4}\)
Vậy \(x=\frac{3}{4}\)
\(\left(\frac{1}{3}x-\frac{1}{4}\right)^2+\left(x^2-\frac{9}{16}\right)^4=0\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{3}x-\frac{1}{4}=0\\x^2-\frac{9}{16}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{3}x=\frac{1}{4}\\x^2=\frac{9}{16}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\end{cases}\Rightarrow}x=\frac{3}{4}}\)\(\Rightarrow\hept{\begin{cases}\frac{1}{3}x-\frac{1}{4}=0\\x^2-\frac{9}{16}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{3}x=\frac{1}{4}\\x^2=\frac{9}{16}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\x=\pm\frac{3}{4}\end{cases}\Rightarrow}x=\frac{3}{4}}\)
Vậy x=3/4
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