$a)\dfrac{3{{x}^{2}}+7x-10}{x}=0$
ĐK: $x\ne 0$
$\begin{align}
& Pt\Leftrightarrow 3{{x}^{2}}-3x+10x-10=0 \\
& \Leftrightarrow 3x\left( x-1 \right)+10\left( x-1 \right)=0 \\
& \Leftrightarrow \left( x-1 \right)\left( 3x+10 \right)=0 \\
& \Leftrightarrow \left[ \begin{align}
& x-1=0 \\
& 3x+10=0 \\
\end{align} \right.\Leftrightarrow \left[ \begin{align}
& x=1 \\
& x=-\dfrac{10}{3} \\
\end{align} \right.\left( tm \right) \\
\end{align}$
$b)\dfrac{4x-17}{2{{x}^{2}}+1}=0$
ĐK: $x\in \mathbb{R}$
$Pt\Leftrightarrow 4x-17=0\Rightarrow x=\dfrac{17}{4}\left( tm \right)$
a. ĐK: x\(\ne\)0
Để phương trình trên bằng 0 thì 3x2+7x-10=0 \(\Leftrightarrow\left[{}\begin{matrix}x=1\:\left(TM\right)\\x=-\frac{10}{3}\left(TM\right)\end{matrix}\right.\)
Vậy...
b. ĐK: 2x2+1\(\ne\)0 \(\forall\)x
Để phương trình trên bằng 0 thì 4x-17=0 \(\Leftrightarrow x=\frac{17}{4}\)(TM)
Vậy...
c. ĐK: x+5\(\ne\)0 \(\Leftrightarrow\)x\(\ne\)-5
Ta có: \(\frac{2x-5}{x+5}=3\Leftrightarrow2x-5=3\left(x+5\right)\)
\(\Leftrightarrow x=-20\)(TM)
Vậy...
d. ĐK: 3x+2\(\ne\)0 \(\Leftrightarrow x\text{}\ne-\frac{2}{3}\)
Ta có: \(\frac{5}{3x+2}=2x-1\Leftrightarrow5=\left(2x-1\right)\left(3x+2\right)\)
\(\Leftrightarrow6x^2+4x-3x-2=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x=-\frac{7}{6}\left(TM\right)\end{matrix}\right.\)
Vậy...
$c)\dfrac{2x-5}{x+5}=3$
ĐK: $x\ne -5$
$Pt\Leftrightarrow 2x-5=3\left( x+5 \right)$
$\Leftrightarrow 2x-5=3x+15$
$\Leftrightarrow 2x-3x=15+5$
$\Leftrightarrow x=-20\left( tm \right)$
$d)\dfrac{5}{3x+2}=2x-1$
ĐK: $x\ne -\dfrac{2}{3}$
$Pt\Leftrightarrow 5=\left( 2x-1 \right)\left( 3x+2 \right)$
$\Leftrightarrow 5=6{{x}^{2}}+x-2$
$\Leftrightarrow 6{{x}^{2}}+x-7=0$
$\Leftrightarrow \left( x-1 \right)\left( 6x+7 \right)=0$
$\Leftrightarrow \left[ \begin{align}
& x-1=0 \\
& 6x+7=0 \\
\end{align} \right.\Leftrightarrow \left[ \begin{align}
& x=1 \\
& x=-\dfrac{7}{6} \\
\end{align} \right.\left( tm \right)$
