\(\left(x^2-16\right)^2-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(x^2-16-x+4\right)\left(x^2-16+x-4\right)=0\)
\(\Leftrightarrow\left(x^2-x-12\right)\left(x^2+x-20\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)\left(x-4\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\\x=-5\end{matrix}\right.\)
\(\left(x^2-16\right)^2-\left(x-4\right)^2\)
\(=\left[\left(x-4\right)\left(x+4\right)\right]^2-\left(x-4\right)^2\)
\(=\left(x-4\right)^2\left[\left(x+4\right)^2-1\right]\)
\(=\)\(\left(x-4\right)^2\left(x^2+8x+16-1\right)\)
\(=\left(x-4\right)^2\left(x^2+8x+15\right)\)
\(=\left(x-4\right)^2\left[\left(x^2+3x\right)+\left(5x+15\right)\right]\)
\(=\left(x-4\right)^2+\left[x\left(x+3\right)+5\left(x+3\right)\right]\)
\(=\left(x-4\right)^2\left(x+5\right)\left(x+3\right)\)
làm tiếp nhé, quên bỏ dở giữa chừng rồi
(x2−16)2−(x−4)2
=[(x−4)(x+4)]2−(x−4)2
=(x−4)2[(x+4)2−1]
=(x−4)2(x2+8x+16−1)
=(x−4)2(x2+8x+15)
=(x−4)2[(x2+3x)+(5x+15)]
=(x−4)2+[x(x+3)+5(x+3)]
