Bài 1:
a) \(\left(x+2\right)\left(3x+5\right)-5x\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+5-5x\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy phương trình đã cho có 2 nghiệm là : \(x_1=-2;x_2=\dfrac{5}{2}\)
b) \(\dfrac{5x-1}{3}=\dfrac{2x-3}{5}-1\)
\(\Leftrightarrow5\left(5x-1\right)=3\left(2x-3\right)-15\)
\(\Leftrightarrow25x-5=6x-9-15\)
\(\Leftrightarrow25x-6x=-9-15+5\)
\(\Leftrightarrow19x=-19\)
\(\Leftrightarrow x=-1\)
Vậy phương trình đã cho có 1 nghiệm là : x=-1
c) \(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(1-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy phương trình đã cho có 2 nghiệm là : \(x_1=-2;x_2=\dfrac{1}{5}\)
d) \(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{3x^2+5x}{x^2-25}\left(ĐKXĐ:x\ne\pm5\right)\)
\(\Rightarrow\left(x+5\right)^2-\left(x-5\right)^2=3x^2+5x\)
\(\Leftrightarrow\left[\left(x+5\right)-\left(x-5\right)\right].\left[\left(x+5\right)+\left(x-5\right)\right]=3x^2+5x\)
\(\Leftrightarrow\left(x+5-x+5\right)\left(x+5+x-5\right)=3x^2+5x\)
\(\Leftrightarrow10.2x=3x^2+5x\)
\(\Leftrightarrow20x-3x^2-5x=0\)
\(\Leftrightarrow3x^2-15x=0\)
\(\Leftrightarrow3x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
vậy phương trình có 2 nghiệm là : \(x_1=0;x_2=5\)
