a) \(x^2+2x=\left(x-2\right).3x\)
\(\Leftrightarrow x^2+2x=3x^2-6x\)
\(\Leftrightarrow x^2+2x-3x^2+6x=0\)
\(\Leftrightarrow-2x^2+8x=0\)
\(\Leftrightarrow-2x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy S = {0;4}
b) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\mp1\end{matrix}\right.\)
Vậy: S = {-1; 1}
c) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]=40\)
\(\Leftrightarrow\left(x^2+5x+x+5\right)\left(x^2+4x+2x+8\right)=40\)
\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt x2 + 6x + 5 = t
\(\Leftrightarrow t.\left(t+3\right)=40\)
\(\Leftrightarrow t^2+3t=40\)
\(\Leftrightarrow t^2+2.t.\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{169}{4}\)
\(\Leftrightarrow\left(t+\dfrac{3}{2}\right)^2=\dfrac{169}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}t+\dfrac{3}{2}=\dfrac{13}{2}\\t+\dfrac{3}{2}=-\dfrac{13}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{13}{2}-\dfrac{3}{2}=\dfrac{10}{2}=5\\t=-\dfrac{13}{2}-\dfrac{3}{2}=-\dfrac{16}{2}=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x+5=5\\x^2+6x+5=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\)
Mà: \(x^2+6x+13=x^2+2.x.3+9+4=\left(x+3\right)^2+4\ne0\)
=> x2 + 6x = 0
<=> x. (x + 6) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy S = {0; -6}
a) Ta có: \(x^2+2x=\left(x-2\right)\cdot3x\)
\(\Leftrightarrow x\left(x+2\right)-3x\left(x-2\right)=0\)
\(\Leftrightarrow x\left[\left(x+2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x+2-3x+6\right)=0\)
\(\Leftrightarrow x\left(-2x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy: S={0;4}
b) Ta có: \(x^3+x^2-x-1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(x-1\right)\cdot\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy: S={-1;1}
c) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-40=0\)
\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)
\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+40-40=0\)
\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)=0\)
\(\Leftrightarrow\left(x^2+6x\right)\left(x^2+6x+13\right)=0\)
\(\Leftrightarrow x\left(x+6\right)\left(x^2+6x+13\right)=0\)
mà \(x^2+6x+13>0\forall x\)
nên \(x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy: S={0;-6}
