(a - b)(a² + 3ab + b²) + (a + b)³ + ab(b - a)

= (a - b)(a² + 3ab + b²) + (a + b)(a² + 2ab + b²) + ab(b - a)

= a(a² + 3ab + b²) - b(a² + 3ab + b²) + a(a² + 2ab + b²) + b(a² + 2ab + b²) + ab(b - a)

= a³ + 3a²b + ab² - ba² - 3ab² - b³ + a³ + 2a²b + ab² + ba² + 2ab² + b³ + ab² - a²b

= 2a³ + 4a²b + 2ab²

Gọi I là trung điểm của AB
=> EI song song MB
=> ^AEI = ^AMB = 60 độ
Do đó ta sẽ chứng minh : ΔEID = ΔEAF 
thì khi đó : ^AEI = ^FED = 60 độ
Thật vậy : EI = 1/2 MB = AE,ID = 1/2 AC = AF
Lại có : ^EAF = 360 − 60.2 − ^BAC = 240 − ^BAC
             ^EID = 360 − 120 − ^BID = 240 − ^BAC
Do đó : ΔEID = ΔEAF (c.g.c)
Tương tự thì : ^EFD=60 độ 
=> đpcm

Bạn giải thích rõ hơn chỗ chứng minh góc EID với góc EAF bằng nhau được không

a,5x^2 - 10xy + 5y^2 - 20z^2
=5(x^2 -2xy +y^2-4z^2 )
=5[(x-y)^2-(2z)^2 ]
=5 .(x-y-2z)(x-y+2z)

b,.= (5x^2+5xy)-(x+y)
=5x(x+y)-(x+y)
=(x+y)(5x-1)

d,x² - 4x + 3 = x² - x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)

e,x² - x - 6 = x² +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)

f,x² - x - 6 = x² +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)

g,2x^2(3x - 5)
= 2x^2 x 3x - 2x^2 x 5
= 6x^3 - 10x^2

\(\text{1) }\)

\(\text{a) }5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)

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

\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)

\(\text{b) }5x^2+5xy-x-y\)

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

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

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

\(\text{c) }2\left(x+4\right)-x^2+16\)

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

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

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

\(=\left(x+4\right)\left(6-x\right)\)

\(\text{d) }x^2+4x+3\)

\(=x^2+3x+x+3\)

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

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

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

\(\text{e) }x^2+5x-6\)

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

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

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

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

Trả lời

P=(a+b+c)³-(a+b-c)³-(b+c-a)³-(c+a-b)³

Đặt a+b-c=x,    b+c-a=y,    c+a-b=z

=>(a+b+c)³-x³-y³-z³

Có x+y+z=a+b-c+b+c-a+c+a-b=a+b+c

=>(x+y+z)³-x³-y³-z³

=>[ (x+y)+z³ ]-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)

Áp dụng hằng đẳng thức trên ta có:

3(a+b-c+b+c-a) (b+c-a+c+a-b) (a+b-c+c+a-b)

=3.2b.2c.2a

=24abc

mk sẽ chỉ hướng để bạn làm bài

đầu tiên ta sẽ nhóm [ (a+b+c)³-(a+b+c)³ ]   ở đây ta thấy có hằng đẳng thức

                                 - [ (b+c-a)³ + ( c+a-b)³ ]    đây cũng vậy 

                sau khi khai triển ta sẽ rút gọn sẽ có nhân tử là  2c 

\(\text{Có: }x^2+y^2+z^2=xy+yz+xz\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)=2\left(xy+yz+xz\right)\)

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

\(\Leftrightarrow x^2+x^2+y^2+y^2+z^2+z^2-2xy-2yz-2xz=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

\(\text{Vì }\left(x-y\right)^2\ge0;\left(y-z\right)^2\ge0\text{ và }\left(x-z\right)^2\ge0\)

\(\text{Nên để }\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

\(\text{thì }\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(x-z\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}\Leftrightarrow}x=y=z}\)

\(\text{Khi đó: }x^{2011}+y^{2011}+z^{2011}=3^{2012}\)

\(\Leftrightarrow x^{2011}+x^{2011}+x^{2011}=3^{2012}\left(\text{Vì x = y = z}\right)\)

\(\Leftrightarrow3x^{2011}=3^{2012}\)

\(\Leftrightarrow x^{2011}=3^{2011}\)

\(\Leftrightarrow x=3\)

\(\text{Vậy }x=y=z=3\)

x⁴+2012x²+2011x+2012

=(x⁴-x)+(2012x²+2012x+2012)

=x(x³-1)+2012(x²+x+1)

=x(x-1) (x²+x+1) + 2012 (x²+x+1)

=(x²+x+1) [x(x-1)+2012]

=(x²+x+1) (x²-x+2012)

\(x^4+2012x^2+2011x+2012\)

\(=x^4-x+2012x^2+2012x+2012\)

\(=x.\left(x-1\right)\left(x^2+x+1\right)+2012.\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2012\right)\)

     \(\left(x^2+3x+1\right)^2-1^2\)

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

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

    \(\left(x^2+3x\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)

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

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