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H
21 tháng 2 2023
\(\dfrac{8}{x+5}=\dfrac{7}{x-4}\) 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)`
NH
21 tháng 2 2023
\(\dfrac{8}{x+5}=\dfrac{7}{x-4}\left(x\ne-5;x\ne4\right)\)
suy ra: \(8\left(x-4\right)=7\left(x+5\right)\\ < =>8x-32=7x+35\\ < =>8x-7x=35+32\\ < =>x=67\left(tm\right)\)
H
21 tháng 2 2023
\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\\ ĐK:\left\{{}\begin{matrix}x\ne-1\\x\ne5\end{matrix}\right.\\ \Leftrightarrow2.\left(x-5\right)=4.\left(x+1\right)\\ \Leftrightarrow2x-10=4x+4\\ \Leftrightarrow2x-4x=4+10\\ \Leftrightarrow-2x=14\\ \Leftrightarrow x=-7\left(t/m\right)\)
NH
21 tháng 2 2023
\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\left(x\ne-1;x\ne5\right)\)
suy ra: \(2\left(x-5\right)=4\left(x+1\right)\\ < =>2x-10=4x+4\\ < =>2x-4x=4+10\\< =>-2x=14\\ < =>x=-7\)
15 tháng 3 2021
Ta có: \(2x+\dfrac{3}{4}=1+\dfrac{x-7}{5}\)
\(\Leftrightarrow\dfrac{40x}{20}+\dfrac{15}{20}=\dfrac{20}{20}+\dfrac{4\left(x-7\right)}{20}\)
\(\Leftrightarrow40x+15=20+4x-28\)
\(\Leftrightarrow40x+15-4x+8=0\)
\(\Leftrightarrow36x+23=0\)
\(\Leftrightarrow36x=-23\)
\(\Leftrightarrow x=-\dfrac{23}{36}\)
Vậy: \(S=\left\{-\dfrac{23}{36}\right\}\)
LN
0
TN
9 tháng 3 2020
\(a.\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\\\Leftrightarrow \frac{3\left(x-5\right)}{12}-\frac{24}{12}x+\frac{12}{12}=\frac{4x}{12}-\frac{2\left(2-x\right)}{12}\\\Leftrightarrow 3\left(x-5\right)-24x+12=4x-2\left(2-x\right)\\\Leftrightarrow 3x-15-24x+12=4x-4+2x\\ \Leftrightarrow3x-15-24x+12-4x+4-2x=0\\ \Leftrightarrow-27x+1=0\\ \Leftrightarrow-27x=-1\\ \Leftrightarrow x=\frac{1}{27}\)
\(b.\left(2x-1\right)^2=\left(x-2\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
\(c.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{-3}{25-x^2}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{x^2-25}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=3\\\Leftrightarrow x^2+5x+5x+25-\left(x^2-5x-5x+25\right)=3\\\Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25=3\\ \Leftrightarrow20x=3\\ \Leftrightarrow x=\frac{3}{20}\)
\(d.x^2-x-12=0\\\Leftrightarrow x^2-4x+3x-12=0\\\Leftrightarrow \left(x^2-4x\right)+\left(3x-12\right)=0\\ \Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
PD
1
\(\dfrac{7}{x-1}=\dfrac{5}{x-4}\left(x\ne1;4\right)\)
\(\Leftrightarrow\dfrac{7\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}=\dfrac{5\left(x-1\right)}{\left(x-1\right)\left(x-4\right)}\)
\(\Leftrightarrow7x-28=5x-5\)
\(\Leftrightarrow7x-5x=-5+28\)
\(\Leftrightarrow2x=23\)
\(\Leftrightarrow x=\dfrac{23}{2}\left(Thoaman\right)\)
\(\dfrac{7}{x-1}=\dfrac{5}{x-4}\)
\(\Leftrightarrow7\left(x-4\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow7x-28-5x+5=0\)
\(\Leftrightarrow2x=23\)
\(\Leftrightarrow x=\dfrac{23}{2}\)