Giải phương trình:
ĐKXĐ : `x \ne -1 ; x \ne 0`
`<=> ( x + 3 )/( x + 1 ) - (( 3x+1)(x+1))/(x(x+1)=(x-1)/x`
`<=> ( x + 3 )/( x + 1 ) - ( 3x + 1 )/( x ) = ( x-1)/x`
`<=> (x+3)/(x+1)=(4x)/x`
`<=> (x+3)/(x+1)=4`
`<=> x+3=4x+4`
`<=> 3x+1=0`
`<=> x=-1/3` ( tm )
Vậy pt có tập nghiệm `S={-1/3}`
\(đk:\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\\ \Leftrightarrow\dfrac{\left(x+3\right)x-3x^2-4x-1}{x\left(x+1\right)}=\dfrac{x^2-1}{x\left(x+1\right)}\\ 0\Leftrightarrow x^2+3x-3x^2-4x-1=x^2-1\\ \Leftrightarrow-2x^2-x-1-x^2+1=0\\ \Rightarrow-3x^2-x=0\\ \Leftrightarrow-x\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(kot/m\right)\\x=-\dfrac{1}{3}\left(t/m\right)\end{matrix}\right.\)
\(\dfrac{x+3}{x+1}-\dfrac{3x^2+4x+1}{x\left(x+1\right)}=\dfrac{x-1}{x}\left(dkxd:x\ne0;-1\right)\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)-3x^2-4x-1-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}=0\)
\(\Leftrightarrow x^2+3x-3x^2-4x-1-x^2+1=0\)
\(\Leftrightarrow-3x^2-x=0\)
\(\Leftrightarrow x\left(-3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\-3x-1=0\end{matrix}\right.\)
\(\Leftrightarrow-3x=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\left(n\right)\)
Vậy \(S=\left\{-\dfrac{1}{3}\right\}\)
`[x+3]/[x+1]-[3x^2+4x+1]/[x(x+1)]=[x-1]/x` `ĐK: x ne 0,x ne -1`
`=>x(x+3)-3x^2-4x-1=(x-1)(x+1)`
`<=>x^2+3x-3x^2-4x-1=x^2-1`
`<=>3x^2+x=0`
`<=>x(3x+1)=0`
`<=>[(x=0(ko t//m)),(x=-1/3(t//m)):}`
b. ĐK :\(x\ne-1\\;x\ne0\)
\(\dfrac{x+3}{x+1}-\dfrac{3x^2+4x+1}{x\left(x+1\right)}=\dfrac{x-1}{x}\\ \Leftrightarrow\dfrac{x^2+3x-3x^2-4x-1}{x\left(x+1\right)}=\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\\ -2x^2-x-1=x^2-1\\ \Leftrightarrow3x^2+x=0\\ x\left(3x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\left(KTM\right)\\x=-\dfrac{1}{3}\left(TM\right)\end{matrix}\right.\)
\(\Rightarrow x=-\dfrac{1}{3}\)
