x+5/x-4
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Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\Rightarrow x=3k;y=5k\)
\(x^2-y^2=-4\\ \Rightarrow9k^2-25k^2=-4\\ \Rightarrow-16k^2=-4\Rightarrow k^2=4\\ \Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6;y=10\\x=-6;y=-10\end{matrix}\right.\)
231+(312-x)=531
<=> 312 - x = 531 - 231
<=> 312 - x = 300
<=> - x = 300 - 312
<=> - x = - 12
<=> x = 12
\(a,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.nghiệm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{1}{6}\)
\(b,\Leftrightarrow\left|x\right|=\dfrac{3}{4}+x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}+x\left(x\ge0\right)\\x=-\dfrac{3}{4}-x\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=\dfrac{3}{4}\left(vô.nghiệm\right)\\x=-\dfrac{3}{8}\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow x=-\dfrac{3}{8}\)
\(\Leftrightarrow164-4\left(x-5\right)=80\\ \Leftrightarrow4\left(x-5\right)=84\\ \Leftrightarrow x-5=21\Leftrightarrow x=26\)
a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)
\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)
\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)
\(\dfrac{3}{2}\)(\(x\) - \(\dfrac{5}{3}\)) - \(\dfrac{4}{5}\) = \(x\) + 1
\(\dfrac{3}{2}\) \(x\) - \(\dfrac{15}{6}\) - \(\dfrac{4}{5}\) = \(x\) + 1
\(\dfrac{3}{2}\)\(x\) - \(x\) = 1 + \(\dfrac{15}{6}\) + \(\dfrac{4}{5}\)
\(\dfrac{1}{2}\)\(x\) =\(\dfrac{43}{10}\)
\(x\) = \(\dfrac{43}{10}\) \(\times\) 2
\(x\) = \(\dfrac{43}{5}\)
\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3.\left(x-\dfrac{5}{3}\right)}{2}-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3x-5}{2}-\dfrac{4}{5}=x+1\Rightarrow\dfrac{5\left(3x-5\right)}{10}-\dfrac{8}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}-x=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1-x\\ \Rightarrow5x-33=10\\ \Rightarrow5x=10+33\\\Rightarrow5x=43\\ \Rightarrow x=\dfrac{43}{5} \)