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\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
a) Quy đồng bỏ mẫu rồi giai pt ta đc : \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
b)\(x=1\)
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
\(5x-200=\frac{5x}{2}-300+2x+300\)0
\(3x-2,5x=200\)\(0,5x=200\)\(x=400\)
Bài 6:
1) Ta có: \(2x\left(x-5\right)-\left(x+3\right)^2=3x-x\left(5-x\right)\)
\(\Leftrightarrow2x^2-10x-\left(x^2+6x+9\right)=3x-5x+x^2\)
\(\Leftrightarrow2x^2-10x-x^2-6x-9-3x+5x-x^2=0\)
\(\Leftrightarrow-14x-9=0\)
\(\Leftrightarrow-14x=9\)
\(\Leftrightarrow x=-\dfrac{9}{14}\)
Vậy: \(S=\left\{-\dfrac{9}{14}\right\}\)
`1)2x(x-5)-(x+3)^2=3x-x(5-x)`
`<=>2x^2-10x-x^2-6x-9=3x-5x+x^2`
`<=>x^2-16x-9=x^2-2x`
`<=>14x=-9`
`<=>x=-9/14`
