\(x^5+x^4+1\)

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

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

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

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

cảm ơn bạn nhiều, không biết còn cách không? Mong nhận đượ giúp đỡ!

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

= \(16-4x+x^2+x^2-5x-4x+20-4\left(x^2-5x+25\right)+1\)

= \(37-13x+2x^2-4x^2+20x+100\)

= \(137+7x-2x^2\)

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

\(=\left(x-4\right)\left(2x-9\right)-\left(\left(2x-10\right)^2-1\right)\)

\(=\left(x-4\right)\left(2x-9\right)-\left(2x-11\right)\left(2x-9\right)\)

\(=\left(2x-9\right)\left(x-4-2x+11\right)=\left(2x-9\right)\left(7-x\right)\)

Lời giải:

1. 
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$

$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$

2.

$4x(x-3y)+12y(3y-x)=4x(x-3y)-12y(x-3y)=(x-3y)(4x-12y)=4(x-3y)(x-3y)=4(x-3y)^2$

3.

$x^3+2x^2-2x-1=(x^3-x^2)+(3x^2-3x)+(x-1)=x^2(x-1)+3x(x-1)+(x-1)$

$=(x-1)(x^2+3x+1)$

1) x²- 3x - 6x +18 

= (x²- 3x )-(6x -18 )

= x(x-3)- 6(x-3)

= (x-6)(x-3)

\(2,x^2-3x-54\)

\(=x^2-9x+6x-54\)

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

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

\(3,20x^2+7x-6\)

\(=20x^2-8x+15x-6\)

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

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

a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz

= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]

= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)

= (xy + yz + zx)(x + y + z)

b) Vô câu hỏi tương tự 

a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz

= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]

= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)

= (xy + yz + zx)(x + y + z)

b) tương tự 

1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

x^2 + 2y^2 - 2y - 2xy + 1  = (x^2 - 2xy + y^2) + (y^2 - 2y + 1) = (x - y)^2 + (y - 1)^2

\(x^2+2y^2-2y-2xy+1\)

\(=x^2-2xy+y^2+y^2-2y+1\)

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

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

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

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

bạn ơi hình như sai đề thì phải a bạn mình nghĩ phải là \(\left(x^2-x+2\right)^2\)

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

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

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

\(=2x^2+2-x=2x^2+2.1-x=2\left(x^2+1\right)-x\)

... Tớ chịu cậu thật đấy !

\(x^2-2^2\) phân tích kiểu chó j đấy >?

\(\left(x-2\right)^2=x^2-2x2+2^2\)

Thôi thôi, bn giải nghệ đi lak vừa.