a: =>x^2+2x+1=x^2+4x-5
=>2x+1=4x-5
=>-2x=-6
=>x=3
b: =>(x+1)(x+3)=0
=>x=-1 hoặc x=-3
c: =>(3x-1)(x+3)-(2x+5)(x-1)+4=x^2+2x-3
=>3x^2+9x-x-3-2x^2+2x-5x+5+4=x^2+2x-3
=>x^2-5x+6=x^2+2x-3
=>-7x=-9
=>x=9/7
a
<=> \(x^2+2x+1=x^2+5x-x-5\)
<=> \(x^2+2x+1-x^2-4x+5=0\)
<=> \(-2x=-6\)
\(\Rightarrow x=3\)
b
<=> \(x^2+3x+x+3=0\)
<=> \(x\left(x+3\right)+\left(x+3\right)=0\)
<=> \(\left(x+1\right)\left(x+3\right)=0\)
=> \(\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
c
ĐK: \(\left[{}\begin{matrix}x-1\ne0\\x+3\ne0\\x^2+2x-3\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-3\\x^2-x+3x-3\ne0\Leftrightarrow\left(x+3\right)\left(x-1\right)\ne0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ne1\\x\ne-3\end{matrix}\right.\)
Phương trình tương đương với:
\(\dfrac{\left(3x-1\right)\left(x+3\right)}{x^2+2x-3}-\dfrac{\left(2x+5\right)\left(x-1\right)}{x^2+2x-3}+\dfrac{4}{x^2+2x-3}-\dfrac{x^2+2x-3}{x^2+2x-3}=0\)
<=> \(\dfrac{\left(3x^2+9x-x-3\right)-\left(2x^2-2x+5x-5\right)+4-x^2-2x+3}{x^2+2x-3}=0\)
<=> \(3x^2+8x-3-2x^2+2x-5x+5+4-x^2-2x+3=0\)
<=> \(3x+9=0\Rightarrow x=-3\left(loại\right)\)
Vậy phương trình vô nghiệm
