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\(\dfrac{5x+3}{x-4}=2\) (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 \(S=\left\{\dfrac{-11}{3}\right\}\)
\(\dfrac{5x+3}{x-4}=2\text{ĐKXĐ:}x\ne4\)
\(\Leftrightarrow\dfrac{5x+3}{x-4}=\dfrac{2\left(x-4\right)}{x-4}MTC:x-4\)
\(\Rightarrow5x+3=2x-8\)
\(\Leftrightarrow5x+3-2x+8=0\)
\(\Leftrightarrow3x+11=0\)
\(\Leftrightarrow3x=-11\)
\(\Leftrightarrow x=\dfrac{-11}{3}\left(\text{nhận}\right)\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-11}{3}\right\}\)
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\(\dfrac{3}{x+5}=\dfrac{4}{x-2}\left(x\ne-5;x\ne2\right)\)
suy ra
`3(x-2)=4(x+5)`
`<=>3x-6=4x+20`
`<=> 3x-4x=20+6`
`<=> -x=26`
`<=> x=-26(tm)`
\(\dfrac{3}{x+5}=\dfrac{4}{x-2}\left(ĐKXĐ:x\ne-5;x\ne2\right)\)
\(\Leftrightarrow3\left(x-2\right)=4\left(x+5\right)\)
\(\Leftrightarrow3x-6=4x+20\)
\(\Leftrightarrow3x-4x=20+6\)
\(\Leftrightarrow-x=26\)
\(\Leftrightarrow x=-26\left(tm\right)\)
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\(\dfrac{2}{x-5}=\dfrac{3}{x+1}\) (ĐK: \(x\ne-1;5\))
\(\Rightarrow2\left(x+1\right)=3\left(x-5\right)\)
\(\Leftrightarrow2x+2=3x-15\)
\(\Leftrightarrow2x-3x=-15-2\)
\(\Leftrightarrow-x=-17\)
\(\Leftrightarrow x=17\) (TMĐK)
\(\dfrac{2}{x-5}=\dfrac{3}{x+1}\text{ĐKXĐ}:x\ne5;-1\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{\left(x-5\right)\left(x+1\right)}=\dfrac{3\left(x-5\right)}{\left(x-5\right)\left(x+1\right)}MTC:\left(x-5\right)\left(x+1\right)\)
\(\Rightarrow2x+2=3x-15\)
\(\Leftrightarrow2x+2-3x+15=0\)
\(\Leftrightarrow17-x=0\)
\(\Leftrightarrow x=17\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{17\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\dfrac{5x+1}{4x-2}+\dfrac{x-3}{x+2}=0\)
\(ĐK:\)
\(\left\{{}\begin{matrix}4x-2\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{1}{2}\\x\ne-2\end{matrix}\right.\)
=> D
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\(\dfrac{3}{x}=\dfrac{5}{x-3}\left(x\ne0;x\ne3\right)\) đề như thế này phải ko ạ?
suy ra: \(3\left(x-3\right)=5x\\ < =>3x-9=5x\\ < =>3x-5x=9\\ < =>-2x=9\\ < =>x=-\dfrac{9}{2}\left(tm\right)\)
\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\left(x\ne-2;x\ne3\right)\)
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`