Bài 5: Phương trình chứa ẩn ở mẫu

\(\dfrac{7x-2}{x+5}=3\) ĐK : \(x\ne-5\)

\(\Leftrightarrow\dfrac{7x-2}{x+5}=\dfrac{3\left(x+5\right)}{x+5}\)

`=>7x-2=3(x+5)`

`<=>7x-2=3x+15`

`<=>7x-3x=15+2`

`<=>4x=17`

`<=>x=17/4(TM)`

`[8x+2]/[x-3]=4`   `ĐK: x ne 3`

`=>8x+2=4(x-3)`

`<=>8x+2=4x-12`

`<=>4x=-14`

`<=>x=-7/2` (t/m)

\(\dfrac{x+2}{x-1}=5\) (ĐK: x ≠ 1)

\(\Leftrightarrow x+2=5\left(x-1\right)\)

\(\Leftrightarrow x+2=5x-5\)

\(\Leftrightarrow x-5x=-5-2\)

\(\Leftrightarrow-4x=-7\)

\(\Leftrightarrow x=\dfrac{7}{4}\) (TMĐK)

\(\dfrac{8}{x+5}=\dfrac{7}{x-4}\) ĐKXĐ : \(\left\{{}\begin{matrix}x+5\ne0\\x-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-5\\x\ne4\end{matrix}\right.\)

`<=> (8(x-4))/((x+5)(x-4)) =(7(x+5))/((x+5)(x-4))`

`=> 8(x-4)=7(x+5)`

`<=>8x-32=7x+35`

`<=>8x-7x=35+32`

`<=>x= 67 (TM)`

\(\dfrac{8}{x+5}=\dfrac{7}{x-4}\left(x\ne-5;x\ne4\right)\)

suy ra: \(8\left(x-4\right)=7\left(x+5\right)\\ < =>8x-32=7x+35\\ < =>8x-7x=35+32\\ < =>x=67\left(tm\right)\)

\(\dfrac{8}{x+5}=\dfrac{7}{x-4}ĐK\left(x\ne5;4\right)\\ 8.\left(x-4\right)=7.\left(x+5\right)\\ 8x-32=7x+35\\ 8x-7x=35+32\\ x=67\)

ĐKXĐ: `x-1\ne0<=>x\ne1` ; `x+4\ne0<=>x\ne-4`

\(\dfrac{7}{x-1}=\dfrac{5}{x+4}\\ \Leftrightarrow7\left(x+4\right)=5\left(x-1\right)\\ \Leftrightarrow7x+28=5x-5\\ \Leftrightarrow7x-5x=-5-28\\ \Leftrightarrow2x=-33\\ \Leftrightarrow x=-\dfrac{33}{2}\left(tm\right)\)

\(\dfrac{4}{x+3}=\dfrac{2}{x-1}\left(x\ne-3;x\ne1\right)\)

suy ra: \(4\left(x-1\right)=2\left(x+3\right)\\ < =>4x-4=2x+6\\ < =>4x-2x=6+4\\ < =>2x=10\\ < =>x=5\left(tm\right)\)

\(\dfrac{4}{x+3}=\dfrac{2}{x-1}\text{ĐKXĐ}:x\ne-3;1\)

\(\Leftrightarrow\dfrac{4\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}MTC:\left(x+3\right)\left(x-1\right)\)

\(\Rightarrow4x-4=2x+6\)

\(\Leftrightarrow4x-4-2x-6=0\)

\(\Leftrightarrow2x-10=0\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\left(\text{nhận}\right)\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{5\right\}\)

ĐKXĐ: `x+3\ne0=>x\ne-3` ; `x-1\ne0=>x\ne1`

\(\dfrac{4}{x+3}=\dfrac{2}{x-1}\\ \Leftrightarrow4\left(x-1\right)=2\left(x+3\right)\\ \Leftrightarrow4x-4=2x+6\\ \Leftrightarrow4x-2x=6+4\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\left(tm\right)\)

\(\dfrac{7}{x-1}=\dfrac{5}{x-4}\left(x\ne1;4\right)\)

\(\Leftrightarrow\dfrac{7\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}=\dfrac{5\left(x-1\right)}{\left(x-1\right)\left(x-4\right)}\)

\(\Leftrightarrow7x-28=5x-5\)

\(\Leftrightarrow7x-5x=-5+28\)

\(\Leftrightarrow2x=23\)

\(\Leftrightarrow x=\dfrac{23}{2}\left(Thoaman\right)\)

\(\dfrac{7}{x-1}=\dfrac{5}{x-4}\)

\(\Leftrightarrow7\left(x-4\right)-5\left(x-1\right)=0\)

\(\Leftrightarrow7x-28-5x+5=0\)

\(\Leftrightarrow2x=23\)

\(\Leftrightarrow x=\dfrac{23}{2}\)

\(\dfrac{7}{x-1}=\dfrac{5}{x-4}\)

\(\Leftrightarrow7\left(x-4\right)=5\left(x-1\right)\)

\(\Leftrightarrow7x-28=5x-5\)

\(\Leftrightarrow7x-5x=-5+28\)

\(\Leftrightarrow2x=23\)

\(\Leftrightarrow x=\dfrac{23}{2}\)

\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\left(x\ne-2;x\ne3\right)\)

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`

\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\text{ĐKXĐ}:x\ne-2;3\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}MTC:\left(x+2\right)\left(x-3\right)\)

\(\Rightarrow3x-9=5x+10\)

\(\Leftrightarrow3x-9-5x-10=0\)

\(\Leftrightarrow-2x-19=0\)

\(\Leftrightarrow-2x=19\)

\(\Leftrightarrow x=\dfrac{-19}{2}\left(\text{nhận}\right)\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-19}{2}\right\}\)

\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\\ ĐK:\left\{{}\begin{matrix}x\ne-1\\x\ne5\end{matrix}\right.\\ \Leftrightarrow2.\left(x-5\right)=4.\left(x+1\right)\\ \Leftrightarrow2x-10=4x+4\\ \Leftrightarrow2x-4x=4+10\\ \Leftrightarrow-2x=14\\ \Leftrightarrow x=-7\left(t/m\right)\)

\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\left(x\ne-1;x\ne5\right)\)

suy ra: \(2\left(x-5\right)=4\left(x+1\right)\\ < =>2x-10=4x+4\\ < =>2x-4x=4+10\\< =>-2x=14\\ < =>x=-7\)

\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\\ \Leftrightarrow2.\left(x-5\right)=4.\left(x+1\right)\\ \Leftrightarrow2x-10=4x+4\\ \Leftrightarrow2x-4x=10+4\\ \Leftrightarrow-2x=14\\ \Leftrightarrow x=14:\left(-2\right)\\ \Leftrightarrow x=-7\)

\(\Leftrightarrow\dfrac{3}{x-4}+\dfrac{4}{x-4}-\dfrac{3x-4}{x^2-16}=0\left(dkxd:x\ne\pm4\right)\)

\(\Leftrightarrow3\left(x+4\right)+4\left(x+4\right)-3x+4=0\)

\(\Leftrightarrow3x+12+4x+16-3x+4=0\)

\(\Leftrightarrow4x+32=0\)

\(\Leftrightarrow x=-8\left(tm\right)\)

Vậy \(S=\left\{-8\right\}\)

\(ĐK:\left\{{}\begin{matrix}x\ne4\\x\ne-4\end{matrix}\right.\\ \Leftrightarrow\dfrac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\\ \Leftrightarrow3\left(x+4\right)-4\left(x-4\right)=3x-4\\ \Leftrightarrow3x+12-4x+16-3x+4=0\\ \Leftrightarrow-4x+32=0\\ \Leftrightarrow-4x=-32\\ \Leftrightarrow x=8\left(t/m\right)\)

\(\dfrac{3}{x-4}-\dfrac{4}{4-x}=\dfrac{3x-4}{x^2-16}\left(Dk:x\ne\pm4\right)\)

\(\Leftrightarrow\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)

\(\Leftrightarrow\dfrac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\dfrac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)

\(\Leftrightarrow3x+12-4x+16=3x-4\)

\(\Leftrightarrow3x-4x-3x=-4-16-12\)

\(\Leftrightarrow-4x=-32\)

\(\Leftrightarrow x=8\left(Thoaman\right)\)