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\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\\ ĐK:\left\{{}\begin{matrix}x\ne-1\\x\ne5\end{matrix}\right.\\ \Leftrightarrow2.\left(x-5\right)=4.\left(x+1\right)\\ \Leftrightarrow2x-10=4x+4\\ \Leftrightarrow2x-4x=4+10\\ \Leftrightarrow-2x=14\\ \Leftrightarrow x=-7\left(t/m\right)\)
\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\left(x\ne-1;x\ne5\right)\)
suy ra: \(2\left(x-5\right)=4\left(x+1\right)\\ < =>2x-10=4x+4\\ < =>2x-4x=4+10\\< =>-2x=14\\ < =>x=-7\)
\(a.\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\\\Leftrightarrow \frac{3\left(x-5\right)}{12}-\frac{24}{12}x+\frac{12}{12}=\frac{4x}{12}-\frac{2\left(2-x\right)}{12}\\\Leftrightarrow 3\left(x-5\right)-24x+12=4x-2\left(2-x\right)\\\Leftrightarrow 3x-15-24x+12=4x-4+2x\\ \Leftrightarrow3x-15-24x+12-4x+4-2x=0\\ \Leftrightarrow-27x+1=0\\ \Leftrightarrow-27x=-1\\ \Leftrightarrow x=\frac{1}{27}\)
\(b.\left(2x-1\right)^2=\left(x-2\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
\(c.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{-3}{25-x^2}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{x^2-25}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=3\\\Leftrightarrow x^2+5x+5x+25-\left(x^2-5x-5x+25\right)=3\\\Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25=3\\ \Leftrightarrow20x=3\\ \Leftrightarrow x=\frac{3}{20}\)
\(d.x^2-x-12=0\\\Leftrightarrow x^2-4x+3x-12=0\\\Leftrightarrow \left(x^2-4x\right)+\left(3x-12\right)=0\\ \Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
\(\dfrac{2}{x-5}=\dfrac{3}{x+1}\) (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)
\(\dfrac{2}{x-5}=\dfrac{3}{x+1}\text{ĐKXĐ}:x\ne5;-1\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{\left(x-5\right)\left(x+1\right)}=\dfrac{3\left(x-5\right)}{\left(x-5\right)\left(x+1\right)}MTC:\left(x-5\right)\left(x+1\right)\)
\(\Rightarrow2x+2=3x-15\)
\(\Leftrightarrow2x+2-3x+15=0\)
\(\Leftrightarrow17-x=0\)
\(\Leftrightarrow x=17\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{17\right\}\)
\(\dfrac{3}{x+5}=\dfrac{4}{x-2}\left(x\ne-5;x\ne2\right)\)
suy ra
`3(x-2)=4(x+5)`
`<=>3x-6=4x+20`
`<=> 3x-4x=20+6`
`<=> -x=26`
`<=> x=-26(tm)`
\(\dfrac{3}{x+5}=\dfrac{4}{x-2}\left(ĐKXĐ:x\ne-5;x\ne2\right)\)
\(\Leftrightarrow3\left(x-2\right)=4\left(x+5\right)\)
\(\Leftrightarrow3x-6=4x+20\)
\(\Leftrightarrow3x-4x=20+6\)
\(\Leftrightarrow-x=26\)
\(\Leftrightarrow x=-26\left(tm\right)\)