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\(\dfrac{8}{x+5}=\dfrac{7}{x-4}\) 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: \(8\left(x-4\right)=7\left(x+5\right)\\ < =>8x-32=7x+35\\ < =>8x-7x=35+32\\ < =>x=67\left(tm\right)\)
\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\\ ĐK:\left\{{}\begin{matrix}x\ne-1\\x\ne5\end{matrix}\right.\\ \Leftrightarrow2.\left(x-5\right)=4.\left(x+1\right)\\ \Leftrightarrow2x-10=4x+4\\ \Leftrightarrow2x-4x=4+10\\ \Leftrightarrow-2x=14\\ \Leftrightarrow x=-7\left(t/m\right)\)
\(\dfrac{2}{x+1}=\dfrac{4}{x-5}\left(x\ne-1;x\ne5\right)\)
suy ra: \(2\left(x-5\right)=4\left(x+1\right)\\ < =>2x-10=4x+4\\ < =>2x-4x=4+10\\< =>-2x=14\\ < =>x=-7\)
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: \(S=\left\{-\dfrac{23}{36}\right\}\)
Bài 1:
a: =>2x-9=10/91
=>2x=829/91
hay x=829/182
b: =>2x=-7
hay x=-7/2
c: =>-3x=-12
hay x=4
ĐKXĐ: `x-1\ne0<=>x\ne1` ; `x+4\ne0<=>x\ne-4`
\(\dfrac{7}{x-1}=\dfrac{5}{x+4}\\ \Leftrightarrow7\left(x+4\right)=5\left(x-1\right)\\ \Leftrightarrow7x+28=5x-5\\ \Leftrightarrow7x-5x=-5-28\\ \Leftrightarrow2x=-33\\ \Leftrightarrow x=-\dfrac{33}{2}\left(tm\right)\)
a: \(\Leftrightarrow\dfrac{y+5}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)
\(\Leftrightarrow2\left(y+5\right)^2-\left(y-5\right)^2=y^2+25y\)
=>\(2y^2+20y+50-y^2+10y-25=y^2+25y\)
=>30y+25=25y
=>5y=-25
=>y=-5(loại)
b: \(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
=>x^2+x+x^2-3x-4x=0
=>2x^2-6x=0
=>2x(x-3)=0
=>x=0(nhận) hoặc x=3(loại)
c: =>x^2-9-6(2x+7)=-13(x+3)
=>x^2-9-12x-42+13x+39=0
=>x^2+x-6=0
=>(x+3)(x-2)=0
=>x=2(nhận) hoặc x=-3(loại)
\(\dfrac{7}{x-1}=\dfrac{5}{x-4}\left(x\ne1;4\right)\)
\(\Leftrightarrow\dfrac{7\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}=\dfrac{5\left(x-1\right)}{\left(x-1\right)\left(x-4\right)}\)
\(\Leftrightarrow7x-28=5x-5\)
\(\Leftrightarrow7x-5x=-5+28\)
\(\Leftrightarrow2x=23\)
\(\Leftrightarrow x=\dfrac{23}{2}\left(Thoaman\right)\)
\(\dfrac{7}{x-1}=\dfrac{5}{x-4}\)
\(\Leftrightarrow7\left(x-4\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow7x-28-5x+5=0\)
\(\Leftrightarrow2x=23\)
\(\Leftrightarrow x=\dfrac{23}{2}\)