a: \(\left(2x^2+x-6\right)+3\left(2x^2+x-3\right)-9=0\)
=>\(2x^2+x-6+6x^2+3x-9-9=0\)
=>\(8x^2+4x-24=0\)
=>\(2x^2+x-6=0\)
=>\(2x^2+4x-3x-6=0\)
=>2x(x+2)-3(x+2)=0
=>(x+2)(2x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(\left|2x-6\right|=3x+4\)
=>\(\left\{{}\begin{matrix}3x+4>=0\\\left(3x+4\right)^2=\left(2x-6\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x>=-4\\\left(3x+4-2x+6\right)\left(3x+4+2x-6\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{4}{3}\\\left(x+10\right)\left(5x-2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{4}{3}\\x\in\left\{-10;\dfrac{2}{5}\right\}\end{matrix}\right.\)
=>\(x=\dfrac{2}{5}\)
c:
ĐKXĐ: \(x\in R\)
Đặt \(x^2+4x+4=a\)
Phương trình sẽ trở thành:
\(\dfrac{a}{a+1}+\dfrac{a+1}{a+2}=\dfrac{7}{6}\)
=>\(\dfrac{a\left(a+2\right)+\left(a+1\right)^2}{\left(a+1\right)\cdot\left(a+2\right)}=\dfrac{7}{6}\)
=>\(7\left(a+1\right)\left(a+2\right)=6\left(a^2+2a+a^2+2a+1\right)\)
=>\(7\left(a^2+3a+2\right)=6\left(2a^2+4a+1\right)\)
=>\(12a^2+24a+6=7a^2+21a+14\)
=>\(5a^2+3a-8=0\)
=>(5a+8)(a-1)=0
=>\(\left[5\left(x^2+4x+4\right)+8\right]\left[x^2+4x+4-1\right]=0\)
=>\(\left[5\left(x+2\right)^2+8\right]\left(x^2+4x+3\right)=0\)
mà \(5\left(x+2\right)^2+8>=8\forall x\)
nên \(x^2+4x+3=0\)
=>(x+1)(x+3)=0
=>\(\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
