CÁCH KHÁC:
\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)
\(<=>x\left(x+10\right)\left(x+4\right)\left(x+6\right)+128\)
\(<=>\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)
\(<=>\left(x^2+10x\right)^2+24\left(x^2+10x\right)+128\)
\(<=>\left(x^2+10x\right)^2+2.\left(x^2+10x\right).12+12^2-16\)
\(<=>\left(x^2+10x+12\right)^2-4^2\)
\(<=>\left(x^2+10x+12-4\right) \left(x^2+10x +12+4\right)\)
\(<=>\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)
\(<=>\left(x^2+10x+8\right)\left(x^2+2x+8x+16\right)\)
\(<=>\left(x^2+10x+8\right)\left[x\left(x+2\right)+8\left(x+2\right)\right]\)
\(<=>\left(x^2+10x+8\right)\left(x+2\right)\left(x+8\right)\)
\(< =>\left[{}\begin{matrix}x^2+10x+8=0\\x+2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-5+\sqrt{17}\\x=-5-\sqrt{17}\\x=-2\\x=-8\end{matrix}\right.\)
Vậy...
Ta có :
\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=0\)
\(\Leftrightarrow\left[x\left(x+10\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+128=0\)
\(\Leftrightarrow\left(x^2+10x\right)\left(x^2+10x+24\right)+128=0\)
\(\Leftrightarrow\left(x^2+10x\right)^2+24\left(x^2+10x\right)+128=0\)
\(\Leftrightarrow\left(x^2+10x\right)^2+24\left(x^2+10x\right)+144=16\)
\(\Leftrightarrow\left(x^2+10x+12\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+10x+12=4\\x^2+10x+12=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2-13=4\\\left(x+5\right)^2-13=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=17\\\left(x+5\right)^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+5=\pm\sqrt{17}\\x+5=\pm3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{17}-5\\\left[{}\begin{matrix}x=-2\\x=-8\end{matrix}\right.\end{matrix}\right.\)
$x(x+4)(x+6)(x+10)+128=[x(x+10)][(x+4)(x+6)]+128$
$<=>(x^2+10x)(x^2+10x+24)+128$
Đặt $x^2+10x+12=y$, đa thức đã cho có dạng
$<=>(y-12)(y+12)+128$
<=>y^2-144+128$
$<=>y^2-16$
$<=>(y+4)(y-4)$
$<=>(x^2+10x+16)(x^2+10x+8)$
$<=>(x^2+2x+8x+16)(x^2+10x+8)$
$<=>(x+2)(x+8)(x^2+10x+8)$
\(< =>\left[{}\begin{matrix}x+2=0\\x+8=0\\x^2+10x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-2\\x=-8\\x=-5+\sqrt{17}\\x=-5-\sqrt{17}\end{matrix}\right.\)
Vậy...
$x(x+4)(x+6)(x+10)+128=[x(x+10)][(x+4)(x+6)]+128$
$<=>(x^2+10x)(x^2+10x+24)+128$
Đặt $x^2+10x+12=y$, đa thức đã cho có dạng
$<=>(y-12)(y+12)+128$
$<=>y^2-144+128$
$<=>y^2-16$
$<=>(y+4)(y-4)$
$<=>(x^2+10x+16)(x^2+10x+8)$
$<=>(x^2+2x+8x+16)(x^2+10x+8)$
$<=>(x+2)(x+8)(x^2+10x+8)$
\(< =>\left[{}\begin{matrix}x+2=0\\x+8=0\\x^2+10x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-2\\x=-8\\x=-5+\sqrt{17}\\x=-5-\sqrt{17}\end{matrix}\right.\)
Vậy...
