k cho tui nha

a) Cách 1.

Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)

= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).

Cách 2.

Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)

= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).

b) Biến đổi được a 4   -   9 rt 3   +   a 2 -9a = (a- 9)a( a 2  +1).

c) Biến đổi được 3 x 2  + 5y - 3xy + (-5x) = (x - y)(3x - 5).

d) Biến đổi được  x 2  - (a + b)x + ab = (x- a)(x - b).

e) Ta có 4 x 2 - 4xy + y 2   –   9 t 2 =  ( 2 x   -   y ) 2   -   ( 3 t ) 2

= (2x - y - 3t )(2x - y + 31).

g) Ta có  x 3   -   3 x 2 y   +   3 xy 2   -   y 3   -   z 3

= ( x   -   y ) 3   -   z 3 = (x - y - z)( x 2   +   y 2   +   z 2  - 2xy + xz - yz).

h) Ta có x 2   -   y 2 + 8x + 6y+ 7 = ( x 2  +8x + 16) - ( y 2  - 6y+ 9)

= ( x   +   4 ) 2   - ( y - 3 ) 2  =(x-y + 7)(x + y + l).

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

\(x^3-x+3x^2+3xy^2+y^3-y\)

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

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

= (x³ + 3x²y + 3xy² + y³) - (x+y)

= (x + y)³ - (x + y)

= (x + y).[(x+y)² - 1 ]

= (x + y).(x + y - 1).(x + y + 1)

a) 12x.                           b) 4xy

c) 2y(3 x 2   +   y 2 ).

d) (x + y + z)( x 2   +   y 2   + z 2  – xy – xz - yz).

1. Ta có: hằng đẳng thức: \(x^3+y^3+z^3=3xyz\) nếu x+y+z=0

đặt b-c=x, c-a=y, a-b=z⇒x+y+z=0

 \(\Rightarrow\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3=3\left(a-b\right)\left(c-a\right)\left(b-c\right)\)

2. \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

3. Tham khảo: https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-y-5-x-5-y-5-thanh-nhan-tu-faq447273.html

\(5,=x^3+2x^2y-7x^2y-14xy^2\\ =x^2\left(x+2y\right)-7xy\left(x+2y\right)\\ =x\left(x-7y\right)\left(x+2y\right)\)

\(\left(x+y-z\right)^3-x^3-y^3+z^3\)

\(=\left[\left(x+y\right)-z\right]^3-x^3-y^3+z^3\)

\(=\left(x+y\right)^3-z^3-3\left(x+y\right)z\left(x+y-z\right)-x^3-y^3+z^3\)

\(=x^3+y^3-z^3+3xy\left(x+y\right)-3\left(x+y\right)z\left(x+y-z\right)-x^3-y^3+z^3\)

\(=3xy\left(x+y\right)-3z\left(x+y\right)\left(x+y-z\right)\)

\(=3\left(x+y\right)\left[xy-z\left(x+y-z\right)\right]\)

\(=3\left(x+y\right)\left(xy-zx-yz+z^2\right)\)

\(=3\left(x+y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]\)

\(=3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)

#\(Urushi\text{☕}\)

Áp dụng (a+b)3 = a3+b3+3ab(a+b), ta có:

(x+y+z)3-x3-y3-z3

=[(x+y)+z]3-x3-y3-z3

=(x+y)3+z3+3z(x+y)(x+y+z)-x3-y3-z3

=x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)-x3-y3-z3

=3(x+y)(xy+xz+yz+z2)

=3(x+y)[x(y+z)+z(y+z)]

=3(x+y)(y+z)(x+z)

=(x+y-z-x)[(x+y-z)^2+x(x+y-z)+x^2]-(y-z)(y^2+yz+z^2)

=(y-z)(x^2+y^2+z^2+2xy-2xz-2yz+x^2+xy-xz+x^2-y^2-yz-z^2)

=(y-z)(3x^2+3xy-3xz-3yz)

=3(y-z)(x^2+xy-xz-yz)

=3(y-z)[x(x+y)-z(x+y)]

=3(y-z)(x+y)(x-z)

a) 16(12 t 2  +1).

b) Gợi ý x 3   +   y 3 = ( x   +   y ) 3  - 3xy(x + y)

(x + y - z)( x 2   +   y 2   +   z 2  - xy + xz + yz).

1)  \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

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

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

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

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

4)  \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

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

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

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

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

nhiều thế

6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)

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

7)   \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)          (NHÂN x + 2 vs x +  5  và  x + 3 vs x + 4 )

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

ĐẶT   \(x^2+7x+11=y\)   ta được :  

\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)

\(=y^2-25=\left(y-5\right)\left(y+5\right)\)

8)  \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)

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

9) sai đề rùi bạn ơi ! đề đúng nè 

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

Ta thấy :  

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

Thay vào biểu thức bài cho ta được : 

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

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

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

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

bài ở trên câu 3 : kết luận là  \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs  

Ta có: ( x - y) z³ + ( y - z ) x³ + ( z - x ) y³  

= ( x - y ) z³ + ( y - z )x³ + ( z - y)y³ + ( y - x ) y³

= ( x - y ) ( z³ - y³ ) + ( y - z ) ( x³ - y³) 

= ( x - y ) ( z - y ) ( z² + zy + y² ) + ( y - z ) ( x - y) ( x² + xy + y² ) 

= ( x - y ) ( y - z ) ( x² + xy + y² - z² - zy - y²) 

= ( x - y ) ( y - z ) [ ( x² - z²) + ( xy - zy) ]

= ( x - y ) ( y - z ) [ ( x - z ) ( x + z ) + y ( x - z ) ]

= ( x - y ) ( y - z ) ( x - z ) ( x + y + z )