\(3x^2-7x+4=\left(3x^2-3x\right)-\left(4x-4\right)=3x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(3x-4\right)\)

3x^2 -7x+4

= 3x^2 -3x-4x+4

= 3x ( x-1) -4(x-1)

= (3x-4)(x-1)

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

\(x^{m+4}-x^{m+3}-x+1=x^{m+3}\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^{m+3}-1\right)\)

Ta có: \(x^{m+4}-x^{m+3}-x+1\)

\(=x^{m+3}\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left(x^{m+3}-1\right)\)

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

x⁴ - x³ - x + 1

= (x⁴ - x³) - (x - 1)

= x³(x - 1) - (x - 1)

= (x³ - 1)(x - 1)

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

\(x^4-2x^3+2x-1\)

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

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

\(x^4+5x^2-6\)

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

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

\(=\left(x^2-1\right)\left(x^2+6\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+6\right)\)

\(x^4+5x^2-6\)

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

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

\(=\left(x^2+6\right)\left(x^2-1^2\right)\)

\(=\left(x^2+6\right)\left(x-1\right)\left(x+1\right)\)

x^5+x^4+1

=x⁵+x⁴+x³+x²+x+1-x³-x²-x

=x³.(x²+x+1)+(x²+x+1)-x.(x²+x+1)

tự xử tiếp

Minh Triều?????

\(4\left(x^2+15x+50\right)\left(x^2+18x+72\right)-3x^2\\ =4\left(x+5\right)\left(x+10\right)\left(x+6\right)\left(x+12\right)-3x^2\\ =4\left(x^2+16x+60\right)\left(x^2+17x+60\right)-3x^2\)

Đặt \(x^2+16x+60=a\)

\(=4a\left(a+x\right)-3x^2\\ =4a^2+4ax-3x^2\\ =\left(2a-x\right)\left(2a+3x\right)\\ =\left[2\left(x^2+16x+60\right)-x\right]\left[2\left(x^2+16x+60\right)+3x\right]\\ =\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)

(x2+15x+50)(x2+18x+72)−3x2=4(x+5)(x+10)(x+6)(x+12)−3x2=4(x2+16x+60)(x2+17x+60)−3x24(�2+15�+50)(�2+18�+72)−3�2=4(�+5)(�+10)(�+6)(�+12)−3�2=4(�2+16�+60)(�2+17�+60)−3�2

Đặt x2+16x+60=a�2+16�+60=�

=4a(a+x)−3x2=4a2+4ax−3x2=(2a−x)(2a+3x)=[2(x2+16x+60)−x][2(x2+16x+60)+3x]=(2x2+31x+120)(2x2+35x+120)

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

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

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

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

\(x^4+2x^3+x^2+x+1\)

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

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

\(5x^2-4\left(x^2-2x+1\right)-5=\left(5x^2-5\right)-4\left(x-1\right)^2=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)

\(= \)\(5x^2-4x^2+8x-4-5\)

\(=\)\(x^2+8x-9\)

\(=x^2+9x-x-9\)

\(=(x-1)(x+9)\)