\(a,\frac{-x}{4}+6=8\)\(\Leftrightarrow\frac{x}{-4}=2\Leftrightarrow x=-8\)
b,\(\frac{-4}{x}-7=-5\Leftrightarrow\frac{-4}{x}=2\Leftrightarrow x=-2\)
c,\(12+\frac{-6}{5x}=17\Leftrightarrow-\frac{6}{5x}=5\Leftrightarrow x=-\frac{6}{25}\)
d,\(\frac{3-x}{7}=\frac{x+5}{4}\Leftrightarrow12-4x=7x+35\Leftrightarrow-11x=23\Leftrightarrow x=-\frac{23}{11}\)
e,\(7-2x=-\frac{3}{3x}=-\frac{1}{x}\Leftrightarrow7x-2x^2+1=0\)
\(\Leftrightarrow-2\left(x^2+\frac{7}{2}x+\frac{49}{16}\right)+\frac{57}{8}=0\Leftrightarrow\left(x+\frac{7}{4}\right)^2=\frac{57}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{7}{4}=\frac{\sqrt{57}}{16}\\x+\frac{7}{4}=-\frac{\sqrt{57}}{16}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{57}-28}{16}\\x=\frac{-\sqrt{57}-28}{16}\end{matrix}\right.\)
a, \(-\frac{x}{4}+6=8\)
=> \(-\frac{x}{4}=8-6=2\)
=> \(x=2.-4=-8\)
Vậy \(x\in\left\{-8\right\}\)
\(b,\frac{4}{-x}-7=-5\)
=> \(\frac{4}{-x}=-5+\left(-7\right)=-12\)
=> \(x=4:12=\frac{1}{3}\)
Vậy \(x\in\left\{\frac{1}{3}\right\}\)
\(c,12+\frac{-6}{5x}=17\)
=> \(-\frac{6}{5x}=17-12=5\)
=> \(5x=-6:5=-\frac{6}{5}\)
=> \(x=-\frac{6}{5}:5=\frac{6}{25}\)
Vậy \(x\in\left\{\frac{6}{25}\right\}\)
\(d,\frac{3-x}{7}=\frac{x+5}{4}\)
=>\(4\left(3-x\right)=7\left(x+5\right)\)
=> \(12-4x=7x+35\)
=> \(-4x-7x=35-12\)
=> \(-11x=23\)
=> \(x=23:\left(-11\right)=-\frac{23}{11}\)
Vậy \(x\in\left\{-\frac{23}{11}\right\}\)
e, \(7-2x=-\frac{3}{3x}\)
=> \(7-2x=-\frac{1}{x}\)
=> \(7=2x+\left(-\frac{1}{x}\right)\)
=> \(7=2x-\frac{1}{x}\)
=> \(7=\frac{2x^2}{x}-\frac{1}{x}\)
=> \(7=\frac{2x^2-1}{x}\)
=> :))