Ta có: \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=55\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=55\)
\(\Leftrightarrow x^4-10x^2-4x^2+40=55\)
\(\Leftrightarrow x^4-14x^2-15=0\)
Đặt \(t=x^2\left(t\ge0\right)\), ta có: \(t^2-14t-15=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=15\\t=-1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{15}\\x=-\sqrt{15}\end{matrix}\right.\)
<=>\(\left(x^2-4\right)\left(x^2-10\right)=55\)
<=>\(x^4-14x^2+40=55\)
<=>\(x^4-14x^2-15=0\)
<=>\(\left(x^2-15\right)\left(x^2+1\right)=0\)
<=>\(x^2-15=0\)(cái kia lun lớn hơn 0)
<=>\(x^2=15\)
<=>\(x=\pm\sqrt{15}\)
\(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=55\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=55\)
\(\Leftrightarrow x^4-10x^2-4x^2+40=55\)
\(\Leftrightarrow x^4-14x^2-15=0\)
\(\Leftrightarrow x^4+x^2-15x^2-15=0\)
\(\Leftrightarrow x^2\left(x^2+1\right)-15\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2-15\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-15=0\\x^2+1=0\left(vô.lí\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{15}\)
(Do lớp 8 chưa hk HĐT chứa căn nên mik ko viết chứ x2 - 15 vẫn là HĐT nhé)
