a) \(=a\left(a^3-9a^2+a-9\right)=a\left[a^2\left(a-9\right)+\left(a-9\right)\right]\)

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

b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

c) \(=x\left(2y+z\right)+3\left(2y+z\right)=\left(2y+z\right)\left(x+3\right)\)

d) \(=x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)\)

\(=\left(x-a\right)\left(x-b\right)\)

a) = a(a³-9a²+a-9)

b) =3x²+5y-3xy-5x

= (3x²-5x)+(5y-3xy)

=x(3x-5)+y(5-3x)

=x(3x-5)-y(3x-5)

=(3x-5)(x-y)

c)2xy +3z+6y+xz

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

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

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

a) \(a^4-9a^3+a^2-9a=a\left(a^3-9a^2+a-9\right)=a\left[\left(a^3-9a^2\right)+\left(a-9\right)\right]=a\left(a^2+1\right)\left(a-9\right)\)

\(a,=a^8-16\\ b,\left(a+c\right)^2-b^2=a^2+2ac+c^2-b^2\\ c,=\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\\ =\left(a^4-b^4\right)\left(a^4+b^4\right)=a^8-b^8\\ d,=\left[\left(3x+y\right)-2\right]^2=\left(3x+y\right)^2-4\left(3x+y\right)+4\\ =9x^2+6xy+y^2-12x-4y+4\\ h,=x^3+64\\ e,=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1=...\\ f,=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\\ =2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\)

e đăng đừng Ctrl+V nhiều quá lóe mắt :vv

\(2,\\ a,\Rightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Rightarrow-12x=24\Rightarrow x=-2\\ b,\Rightarrow x^2-4x+4-x^2-6x-9=45\\ \Rightarrow-10x=50\Rightarrow x=-5\\ c,\Rightarrow x^3-27+4x-x^3=1\\ \Rightarrow4x=28\Rightarrow x=7\\ d,\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\\ \Rightarrow12x=-6\Rightarrow x=-\dfrac{1}{2}\)

\(\left(a^2+4\right)^2-16a^2\\ =\left(a^2+4\right)^2-\left(4a\right)^2\\ =\left(a^2-4a+4\right)\left(a^2+4a+4\right)\\ =\left(a-2\right)^2\left(a+2\right)^2\)

Chọn A.

A

a

\(\)Ta có: \(a+b+c=0 \Rightarrow b+c=-a \Rightarrow (b+c)^2=(-a)^2 \Leftrightarrow b^2+c^2+2bc=a^2 \Leftrightarrow a^2-b^2-c^2=2bc\)

Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)

\(P=...=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)

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Bổ đề \(a+b+c=0 \Leftrightarrow a^3+b^3+c^3\)

Ở đây ta c/m chiều thuận:
Với \(a+b+c=0 \Leftrightarrow a+b=-c \Rightarrow (a+b)^3=(-c)^3 \Leftrightarrow a^3+b^3+3ab(a+b)=-c^3 \Leftrightarrow a^3+b^3+c^3=3abc(QED)\)

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).

Thay \(a=-\left(b+c\right)\) ; \(a+c=-b\) và \(a+b=-c\) vào điều kiện thứ 2 ta có 

\(\left(b+c\right)^2=2\left(-b+1\right)\left(-c-1\right)\)

 <=> \(b^2+c^2+2bc=2bc+2b-2c-2\)

<=> \(\left(b-1\right)^2+\left(c+1\right)^2=0\) <=> \(\left\{{}\begin{matrix}b=1\\c=-1\end{matrix}\right.\)

suy ra: a=0. Vậy A = a² + b² + c² = 2