giai pt giup minh voi a
d) . đkxđ : x khác 0 , x khác -5
\(\Leftrightarrow\left(2x+5\right)\left(x+5\right)=2x.x\)
<=> \(2x^2+10x+5x+25=2x^2\)
<=> \(2x^2+15x+25-2x^2=0\)
<=> \(15x+25=0\)
<=> \(15x=-25\Rightarrow x=\dfrac{-25}{15}=-\dfrac{5}{3}\left(nhận\right)\)
Vậy.....
e).
\(\left|x+2\right|=3x+5\Leftrightarrow\left\{{}\begin{matrix}x+2=3x+5\left(khi\right)x+2\ge0\Leftrightarrow x\ge-2\left(1\right)\\-\left(x+2\right)=3x+5\left(khi\right)x+2< 0\Leftrightarrow x< -2\left(2\right)\end{matrix}\right.\)
Giải pt ( 1) khi \(x\ge-2\) :
\(x+2=3x+5\\ \Leftrightarrow x+2-3x-5=0\\ \Leftrightarrow-2x-3=0\)
<=> \(-2x=3\)
\(\Rightarrow x=\dfrac{-3}{2}\left(nhận\right)\)
Giải pt (2) khi \(x< -2\) :
\(-\left(x+2\right)=3x+5\)
\(\Leftrightarrow-x-2-3x-5=0\)
<=> \(-4x-7=0\)
<=> \(-4x=7\)
<=> \(x=\dfrac{-7}{4}\left(loại\right)\)
Vậy \(S=\left\{-\dfrac{3}{2}\right\}\)
f).
\(\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy ..
\(\dfrac{2x+5}{2x}=\dfrac{x}{x+5}\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
\(\left|x+2\right|=3x+5\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=3x+5\\x+2=-3x-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2-3x-5=0\\x+2+3x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-3=0\\4x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-\dfrac{7}{4}\end{matrix}\right.\)
\(\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
\(a.\dfrac{2x+5}{2x}=\dfrac{x}{x+5}\left(x\ne0;-5\right)\Leftrightarrow\left(2x+5\right)\left(x+5\right)=2x^2\Leftrightarrow2x^2+15x+25=2x^2\Leftrightarrow3x+5=0\Leftrightarrow x=-\dfrac{5}{3}\)
\(b.\left|x+2\right|=3x+5\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+2=3x+5\\x+2=-3x-5\end{matrix}\right.\\x\ge-\dfrac{5}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-\dfrac{7}{4}\end{matrix}\right.\\x\ge-\dfrac{5}{3}\end{matrix}\right.\) \(\Rightarrow x=-\dfrac{3}{2}\)
c. \(\left(2x+1\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
