AI LÀM MÌNH SẼ K ĐÚNG + 1 THẺ KHÓA HỌC T.ANH GIAO TIẾP 

\(a.2x^2+5x+2=2x^2+4x+x+2=\left(x+2\right)\left(2x+1\right)\)

Minh Anh ơi, bạn làm hết được không?

b)=x^3+x^2+4x^2+4x+4x+4=x(x+1)+4x(x+1)+4(x+1)=(x+1)(x+4x+4)

Thiên Ân ơi, bạn giải giúp mình câu c được kh

a)  \(x^3+5x^2+8x+4=x^3+x^2+4x^2+4x+4x+4\)

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

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

b)  \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)

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

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

a, = (x^3-x^2)-(4x^2-4x)+(4x-4)

    = (x-1).(x^2-4x+4) = (x-1).(x-2)^2

b, = (x^3+x^2)-(10x^2+10x)+(16x+16)

 = (x+1).(x^2-10x+16)

 = (x+1).[ (x^2-2x)-(8x-16) ] = (x+1).(x-2).(x-8)

k mk nha

a)= (x^3-x^2)-(4x^2-4x)+(4x-4)

    = (x-1).(x^2-4x+4)

    = (x-1).(x-2)^2

b)= (x^3+x^2)-(10x^2+10x)+(16x+16)

 = (x+1).(x^2-10x+16)

 = (x+1).[ (x^2-2x)-(8x-16) ]

 = (x+1).(x-2).(x-8)

P/s tham khảo nha

x³ + 7x - 6=x² . x + 7x - 2² + 2 = (x² - 2²) + (x+7x)+2=(x-2) . (x+2) + 8x + 2

x³ - 5x + 8x - 4=x² . x -5x + 8x -2² = (x² - 2²) . (x -5x + 8x )=(x-2) . (x+2) . 4x

x³ - 9x² + 6x + 16=x² . x - 9x² + 6x + 16 = (x² - 9x²) . (x+6x) + 16=(x-9x) . (x+9x) . 7x + 16

k mk nha

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

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

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

c) \(x^4-4x^3+8x^2-16x+16=x^4+8x^2+16-\left(4x^3+16x\right)\)

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

b) \(x^6-x^4-9x^3+9x^2=x^4\left(x^2-1\right)-\left(9x^3-9x^2\right)\)

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

\(=\left(x^5+x^4-9x^2\right)\left(x-1\right)=\left(x-1\right)x^2\left(x^3+x^2-9\right)\)

1) \(x^3+2x-3\)

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

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

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

2) \(x^3-6x+4\)

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

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

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

3) \(x^3-2x^2+1\)

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

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

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

4) \(x^3+5x^2-12\)

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

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

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

5) \(x^3-6x^2+9x\) (chắc đề như vậy)

\(=x\left(x^2-6x+9\right)\)

\(=x\left(x-3\right)^2\)

6) \(4x^3-9x^2+5x\)

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

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

\(=x\left[4x\left(x-1\right)-5\left(x-1\right)\right]\)

\(=x\left(x-1\right)\left(4x-5\right)\)

a.

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

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

b.

\(x^3-9x^2+6x+16=\left(x^3-7x^2-8x\right)-\left(2x^2-14x-16\right)\)

\(=x\left(x^2-7x-8\right)-2\left(x^2-7x-8\right)\)

\(=\left(x-2\right)\left(x^2-7x-8\right)=\left(x-2\right)\left(x^2+x-8x-8\right)\)

\(=\left(x-2\right)\left[x\left(x+1\right)-8\left(x+1\right)\right]=\left(x-2\right)\left(x+1\right)\left(x-8\right)\)

c.

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+10+2\right)-24\)

\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+10\right)^2-4\left(x^2+7x+10\right)+6\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+10-4\right)+6\left(x^2+7x+10-4\right)\)

\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)