1)
\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right).\left(x+2\right)\left(x+4\right)-40=0\)
\(\Leftrightarrow\left(x^2+6x+5\right).\left(x^2+6x+8\right)-40=0\)
Đặt \(a=x^2+6x+6\) ta có:
\(\Leftrightarrow\left(a-1\right)\left(a+2\right)-40=0\)
\(\Leftrightarrow a^2+a-2-40=0\)
\(\Leftrightarrow a^2-6x+7x-42=0\)
\(\Leftrightarrow a\left(a-6\right)+7\left(a-6\right)=0\)
\(\Leftrightarrow\left(a-6\right)\left(a+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=6\\a=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x+6=6\\x^2+6x+6=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)
(\(x^2+6x+13=\left(x+3\right)^2+4>0\left(loại\right)\))
Vậy.................
3)
\(\left|x+4\right|=\left|3-2x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=3-2x\\x+4=-3+2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=7\end{matrix}\right.\)
Vậy..........
2)\(x^3-7x^2+15x-25=0\)
\(\Leftrightarrow x^3-5x^2-2x^2+10x+5x-25=0\)
\(\Leftrightarrow x^2\left(x-5\right)-2x\left(x-5\right)+5\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-2x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-1\right)^2+4=0\left(loai\right)\end{matrix}\right.\)
4)\(\left|2x-5\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=2-x\\2x-5=-\left(2-x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5-2+x=0\\2x-5+2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-7=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
S=\(\left\{\dfrac{7}{3};3\right\}\)
