phân tích đa thức thành nhân tử chung ( phương pháp nhóm )

       \(3x^5-10x^4-8x^3-3x^2+10x+8\)

\(=3x^5-3x^4-7x^4+7x^3-15x^3+15x^2-18x^2+18x-8x+8\)

\(=\left(x-1\right)\left(3x^4-7x^3-15x^2-18x-8\right)\)

\(=\left(x-1\right)\left[3x^3\left(x-4\right)+5x^2\left(x-4\right)+5x\left(x-4\right)+2\left(x-4\right)\right]\)

\(=\left(x-1\right)\left(x-4\right)\left[3x^3+5x^2+5x+2\right]\)

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

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

A xác định \(\Leftrightarrow x^2-1\ne0\Leftrightarrow x\ne\left\{1;-1\right\}\)

B xác định \(\Leftrightarrow x^3-8\ne0\Leftrightarrow x^3\ne8\Leftrightarrow x\ne2\)

\(=\left(12x+9x^2+4\right)-\left(6y\right)^2=\left(3x+2\right)^2-\left(6y\right)^2\)

\(=\left(3x+2-6y\right)\left(3x+2+6y\right)\)

k mình cái

12x+9x²+4-36y²

= (9x²+12x+4)-36y²

= (3x+2)²-36y²

= ((3x+2)-6y²)((3x+2)+6y²)

=(3x+2-6y²)(3x+2+6y²)

\(n^4+2n^3-n^2-2n\)

\(=n^3\left(n+2\right)-n\left(n+2\right)\)

\(=\left(n+2\right)\left(n^3-n\right)\)

\(=n\left(n+2\right)\left(n^2-1\right)\)

\(=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

Dễ thấy n-1; n; n+1; n+2 là 4 số liên tiếp => có 2 số chẵn => tích của 4 số chia hết cho 2

=> đpcm

\(\frac{x^4+2x^3+10x-25}{x^2+5}\)

\(=\frac{\left(x^4+5x^2\right)+\left(2x^3+10x\right)-\left(5x^2+25\right)}{x^2+5}\)

\(=\frac{x^2.\left(x^2+5\right)+2x.\left(x^2+5\right)-5.\left(x^2+5\right)}{x^2+5}\)

\(=\frac{\left(x^2+5\right)\left(x^2+2x-5\right)}{x^2+5}\)

\(=x^2+2x-5\)\(\left(x^2+5\ne0\right)\)

Tham khảo nhé~

      \(x^4+2x^3+10x-25\)

\(=x^4+5x^2+2x^3+10x-5x^2-25\)

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

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

Vậy \(\left(x^4+2x^3+10x-25\right):\left(x^2+5\right)=x^2+2x-5\)