\(=\left(4x^2-9\right)^2=\left(2x-3\right)^2\left(2x+3\right)^2\)

\(\left(4x^2-9\right)^2=\left(2x-3\right)^2\left(2x+3\right)^2\)

\(4x^4-16-4x^2-16x\)

\(=4x^2\left(x^2-1\right)-16\left(1+x\right)\)

\(=4x^2\left(x+1\right)\left(x-1\right)-16\left(x+1\right)\)

\(=\left(x+1\right)\left[4x^2\left(x-1\right)-16\right]\)

\(=\left(x+1\right)4\left[x^2\left(x-1\right)-4\right]\)

Nguyễn Văn Tuấn AnhNs r, không biết thì not làm

\(4x^4-16-4x^2-16x\)

\(=4x^2\left(x^2-1\right)-16\left(x+1\right)\)

\(=4x^2\left(x-1\right)\left(x+1\right)-16\left(x+1\right)\)

\(=\left(x+1\right)\left[4x^2\left(x-1\right)-16\right]\)

\(=4\left(x+1\right)\left[x^2\left(x-1\right)-4\right]\)

\(=4\left(x+1\right)\left[x^3-x^2-4\right]\)

\(=4\left(x+1\right)\left[x^3+x^2+2x-2x^2-2x-4\right]\)

\(=4\left(x+1\right)\left[x\left(x^2+x+2\right)-2\left(x^2+x+2\right)\right]\)

\(=4\left(x+1\right)\left(x-2\right)\left(x^2+x+2\right)\)

๖ۣۜMαɾαƙαĭ ⁹⁀ᶦᵈᵒᶫ  bạn đừng ns thế người ta không biết làm nhưng vẫn cố gắng giúp 

nếu bạn ấy làm sai thì vẫn có những người như bạn sửa

https://h7.net/hoi-dap/toan-8/phan-h-da-thuc-x-4-16-thanh-nhan-tu-faq324398.html

\(x^4+16x-12\)

\(=x^4+2x^3-2x^2-2x^3-4x^2+4x+6x^2+12x-12\)

\(=x^2.\left(x^2+2x-2\right)-2x.\left(x^2+2x-2\right)+6.\left(x^2+2x-2\right)\)

\(=\left(x^2+2x-2\right)\left(x^2-2x+6\right)\)

Làm

x⁴  + 16x - 12 

= x⁴  + 2x³ - 2x² - 2x³ - 4x² + 4x + 6x ² + 12x - 12

= x² ( x² + 2x - 2 ) - 2x ( x²  + 2x - 2 ) + 6( x²  + 2x - 2 )

=( x²  + 2x - 2 )( x² - 2x + 6 )

_{HỌC TỐT @}

\(=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)

\(16x-5x^2-3=-\left(5x^2-16x+3\right)=-\left[\left(5x^2-15x\right)-\left(x-3\right)\right]=-\left[5x\left(x-3\right)-\left(x-3\right)\right]=-\left(5x-1\right)\left(x-3\right)=\left(5x-1\right)\left(3-x\right)\)