5x+3phần x-4 bằng 2
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\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\left(x\ne-2;x\ne3\right)\)
suy ra: \(3\left(x-3\right)=5\left(x+2\right)\\ < =>3x-9=5x+10\\ < =>3x-5x=10+9\\ < =>-2x=19\\ < =>x=-\dfrac{19}{2}\left(tm\right)\)
\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\)ĐKXĐ \(\left\{{}\begin{matrix}x+2\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}\)
`<=> 3(x-3) =5 (x+2)`
`<=> 3x-9 = 5x+10`
`<=>3x -5x=10+9`
`<=> -2x=19`
`<=>x=-19/2`
\(\frac{3}{-5}=\frac{-3}{5}\)
\(\Rightarrow\)\(\frac{-3}{5}=\frac{-6}{10}=\frac{-9}{15}=\frac{-15}{25}=\frac{-18}{30}\)
\(\left(\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{17}{4}-\dfrac{3}{4}\)
\(=\left(\dfrac{9}{12}+\dfrac{8}{12}\right):\dfrac{17}{4}-\dfrac{3}{4}\)
\(=\dfrac{17}{12}:\dfrac{17}{4}-\dfrac{3}{4}\)
\(=\dfrac{17.4}{17.12}-\dfrac{3}{4}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}\)
\(=\dfrac{4}{12}-\dfrac{9}{12}\)
\(=-\dfrac{5}{12}\)
\(\dfrac{5x+3}{x-4}=2\) (1)
ĐKXĐ: \(x\ne4\)
(1) \(\Leftrightarrow5x+3=2\left(x-4\right)\)
\(\Leftrightarrow5x+3=2x-8\)
\(\Leftrightarrow5x-2x=-8-3\)
\(\Leftrightarrow3x=-11\)
\(\Leftrightarrow x=\dfrac{-11}{3}\) (nhận)
Vậy \(S=\left\{\dfrac{-11}{3}\right\}\)
\(\dfrac{5x+3}{x-4}=2\text{ĐKXĐ:}x\ne4\)
\(\Leftrightarrow\dfrac{5x+3}{x-4}=\dfrac{2\left(x-4\right)}{x-4}MTC:x-4\)
\(\Rightarrow5x+3=2x-8\)
\(\Leftrightarrow5x+3-2x+8=0\)
\(\Leftrightarrow3x+11=0\)
\(\Leftrightarrow3x=-11\)
\(\Leftrightarrow x=\dfrac{-11}{3}\left(\text{nhận}\right)\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-11}{3}\right\}\)