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

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

\(16x^4+8x^2+1-8x^2\)

\(=\left(4x^2+1\right)^2-8x^2\)

\(=\left(4x^2+1-x\sqrt{8}\right)\left(4x^2+1+x\sqrt{8}\right)\)

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

Phần 1 ) Đánh lộn dấu " ) " thành số 0 rồi bạn .... 

\(\left(x+1\right).\left(x+2\right).\left(x+3\right).\left(x+4\right)-15\)

\(=\left[\left(x+1\right).\left(x+4\right)\right].\left[\left(x+2\right).\left(x+3\right)\right]-15\)

\(=\left(x^2+5x+4\right).\left(x^2+5x+6\right)-15\)

\(\text{Đặt }k=x^2+5x+4,\text{ta có: }\)

\(\left(x^2+5x+4\right).\left(x^2+5x+6\right)-15=k.\left(k+2\right)-15=k^2+2k-15=k^2-5k+3k-15=k.\left(k-5\right)+3.\left(k-5\right)=\left(k-5\right).\left(k+3\right)\)

tự thay vô nha : )

\(16x^3y+\frac{1}{4}yz^3\)

\(\text{Phân tích thành nhân tử}\)

\(\frac{y\left(\frac{z}{2}+2x\right)\left(z^2-4xz+16x^2\right)}{2}\)

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

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

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

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

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

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 @}

\(x^2-4y^2+16x+64=\left(x^2+16x+64\right)-4y^2\)

\(=\left(x+8\right)^2-\left(2y\right)^2\)

\(=\left(x-2y+8\right)\left(x+2y+8\right)\)