Phân tích thành nhân tử

\(=\left(my+nx\right)\left(ny+mx\right)\)

mn(x² +y²) +xy(m² +n²⁾= mnx² +mny² +xym² +xyn²

                                              =mx(nx + my) +ny( my +nx)

                                  =(mx+ny)(nx+my)

Đặt \(x^2+x+1=t\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

= \(\left(x^2+x+1\right)\left[\left(x^2+x+1\right)+1\right]-12\)

= \(\left(x^2+x+1\right)^2\left(x^2+x+1\right)-12\)

= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)

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

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

Ta có \(x^2-\left(m+n\right)x+m.n=\left(x^2-mx\right)-\left(nx-m.n\right)\)

\(=x\left(x-m\right)-n\left(x-m\right)=\left(x-m\right)\left(x-n\right)\)

a: \(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)

\(=a\left(x-y\right)+b\left(x-y\right)+c\left(x-y\right)\)

\(=\left(x-y\right)\left(a+b+c\right)\)

b: \(a^m-a^{m+2}\)

\(=a^m-a^m\cdot a^2\)

\(=a^m\left(1-a^2\right)\)

\(=a^m\left(1-a\right)\left(1+a\right)\)

ta có: \(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)

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

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

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

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

Cho mình nhé hihi!!!

x²(x+4)²-(x+4)²-(x²-1)

=(x+4)²  (x²-1)-(x²-1)

=(x²-1)(x²+8x+16-1)

=(x-1)(x+1)(x²+8x+15)

\(\left(x-1\right)\left(x+1\right)\left(x^2+8x+15\right)\)

x³+27+(x+3)(x+9)

= (x+3)(x²-3x+9)+(x+3)(x+9)

= (x+3)(x²-3x+9+x+9)

=(x+3)(x²-2x+18)

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

x³+27+(x+3)(x+9)

= (x+3)(x²-3x+9)+(x+3)(x+9)

= (x+3)(x²-3x+9+x+9)

=(x+3)(x²-2x+18)

`(x+3)^4+(x+5)^4-2`

`={[(x+3)^2]^2-1^2}+{[(x+5)^2]^2 -1^2}`

`=[(x+3)^2-1^2][(x+3)^2+1]+[(x+5)^2-1^2][(x+5)^2+1]`

`=(x+3-1)(x+3+1)[(x+3)^2+1]+(x+5-1)(x+5+1)[(x+5)^2+1]`

`=(x+2)(x+4)[(x+3)^2+1]+(x+4)(x+6)[(x+5)^2+1]`

`=(x+4){(x+2)[(x+3)^2+1]+(x+6)[(x+5)^2+1]}`

`=(x+4)(2x^3+24x^2+108x+176)`

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

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

\(=\left[\left(x^2+6x+9-1\right)\left(x^2+6x+9+1\right)\right]+\left[\left(x^2+10x+25-1\right)\left(x^2+10x+25+1\right)\right]\)

\(=\left(x^2+6x+8\right)\left(x^2+6x+10\right)+\left(x^2+10x+24\right)\left(x^2+10x+26\right)\)

\(=\left(x+2\right)\left(x+4\right)\left(x^2+6x+10\right)+\left(x+4\right)\left(x+6\right)\left(x^2+10x+26\right)\)

\(=\left(x+4\right)\left[\left(x+2\right)\left(x^2+6x+10\right)+\left(x+6\right)\left(x^2+10x+26\right)\right]\)

\(=\left(x+4\right)\left(x^3+6x^2+10x+2x^2+12x+20+x^3+10x^2+26x+6x^2+60x+156\right)\)

\(=\left(x+4\right)\left(2x^3+24x^2+108x+176\right)\)

\(=2\left(x+4\right)\left(x^3+12x^2+54x+88\right)\)

a, \(x^3-2x-y^3+2y\) (sửa đề)

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

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

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

b, \(\left(x-y\right)\left(x+y\right)-4zx+4yz\)

\(=\left(x-y\right)\left(x+y\right)-\left(4zx-4yz\right)\)

\(=\left(x-y\right)\left(x+y\right)-4z\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-4z\right)\)

Bạn xem lại đề câu a giúp mình nha!

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

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

\(=2\left(x^2+x-5\right)\left(x^2+x-6\right)-3\left(x^2+x-6\right)\)

\(=\left(x^2+x-6\right)\left(2x^2+2x-13\right)\)

\(=\left(x-2\right)\left(x+3\right)\left(2x^2+2x-13\right)\)

\(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)

Đặt t=\(x^2+x\)

\(\Rightarrow C=2\left(t-5\right)^2-5t+28=2t^2-20t+50-5t+28=2t^2-25t+78=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)\)

Thay t: \(C=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)=2\left(x^2+x-\dfrac{13}{2}\right)\left(x^2+x-6\right)=2\left(x-2\right)\left(x+3\right)\left(x^2+x-\dfrac{13}{2}\right)\)

Ta có: \(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)

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

\(=\left(x-2\right)\left(x+3\right)\left(2x^2+2x-13\right)\)