a: \(\left(x+1\right)\left(x+3\right)-x\left(x+2\right)=7\)
=>\(x^2+4x+3-x^2-2x=7\)
=>2x+3=7
=>2x=4
=>x=2
b: 2x(3x+5)-x(6x-1)=33
=>\(6x^2+10x-6x^2+x=33\)
=>11x=33
=>x=3
c: 6x(5x-2)-2(5x-2)=0
=>(5x-2)(6x-2)=0
=>\(\left[{}\begin{matrix}5x-2=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{2}{6}=\dfrac{1}{3}\end{matrix}\right.\)
d: \(\left(x^2+1\right)\left(x-2\right)+2x=4\)
=>\(\left(x^2+1\right)\left(x-2\right)+2x-4=0\)
=>\(\left(x-2\right)\left(x^2+1+2\right)=0\)
mà \(x^2+3>=3>0\forall x\)
nên x-2=0
=>x=2
e:
\(\left(x+1\right)^2=x+1\)
=>\(x^2+2x+1-x-1=0\)
=>\(x^2+x=0\)
=>x(x+1)=0
=>\(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
f: \(\left(x-5\right)^3-x+5=0\)
=>\(\left(x-5\right)^3-\left(x-5\right)=0\)
=>\(\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\)
=>(x-5)(x-5-1)(x-5+1)=0
=>(x-5)(x-6)(x-4)=0
=>\(\left[{}\begin{matrix}x-5=0\\x-6=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
g: \(x^2+5x-6=0\)
=>\(x^2-x+6x-6=0\)
=>x(x-1)+6(x-1)=0
=>(x-1)(x+6)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\)
h: \(x^2-2x-3=0\)
=>\(x^2-3x+x-3=0\)
=>(x-3)(x+1)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
i: \(x^2-x-6=0\)
=>\(x^2-3x+2x-6=0\)
=>x(x-3)+2(x-3)=0
=>(x-3)(x+2)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
k: \(x^3+2x^2-3=0\)
=>\(x^3-x^2+3x^2-3=0\)
=>\(x^2\left(x-1\right)+3\left(x-1\right)\left(x+1\right)=0\)
=>\(\left(x-1\right)\left(x^2+3x+3\right)=0\)
mà \(x^2+3x+3=x^2+3x+\dfrac{9}{4}+\dfrac{3}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
nên x-1=0
=>x=1
l: \(5\left(x-2\right)-x^2+4=0\)
=>\(5\left(x-2\right)-\left(x^2-4\right)=0\)
=>\(\left(x-2\right)\cdot5-\left(x-2\right)\left(x+2\right)=0\)
=>\(\left(x-2\right)\left(5-x-2\right)=0\)
=>(x-2)(3-x)=0
=>\(\left[{}\begin{matrix}x-2=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
