Ta có: x+y+z=0
=>x+y=-z =>(x+y)^5=-z^5
hay x^5+y^5+5(x^4y+xy^4+2x³y²+2x²y³+)=-z^5
<=>x^5+y^5+z^5+5xy(x³+y³+2x²y+2x²y)=0
<=>x5+y^5+z^5+5xy(x+y)(x²-xy+y²+2xy)=0
<=>x^5+y^5+z^5-5xyz(x²+xy+y²)=0
<=>x^5+y^5+z^5=5xyz(x²+xy+y²)
<=>2(x^5+y^5+z^5)=5xyz(2x²+2xy+2y²)
<=>2(x^5+y^5+z^5)=5xyz[x²+y²+(x+y)²]
<=>2(x^5+y^5+z^5)=5xyz(x³+y²+z²)

Từ x+y+z=0 => y+z=-x => (y+z)⁵=-x⁵

=> \(y^5+5y^4z+10y^2z^2+10y^2z^3+5yz^4+z^5=-x^5\)

\(\Rightarrow\left(x^5+y^5+z^5\right)+5yz\left(y^3+2y^2z+2yz^2+z^3\right)=0\)

\(\Rightarrow\left(x^5+y^5+z^5\right)+5yz\left[\left(y+z\right)\left(y^2-yz+x^2\right)\right]=0\)

\(\Rightarrow\left(x^5+y^5+z^5\right)+5yz\left(y+z\right)\left(y^2+yz+z^2\right)=0\)

\(\Rightarrow2\left(x^5+y^5+z^5\right)-5xyz\left[\left(y^2+2yz+z^2\right)+y^2+z^2\right]=0\)

\(\Rightarrow2\left(x^5+y^5+z^5\right)=5xyz\left[\left(y+z\right)^2+y^2+z^2\right]\) (đpcm)

Lời giải:

Ta có:

$x^3+y^3+z^3=(x+y)^3-3xy(x+y)+z^3=(-z)^3-3xy(-z)+z^3$
$=(-z)^3+3xyz+z^3=3xyz$
Khi đó:

$2(x^5+y^5+z^5)=2[(x^3+y^3+z^3)(x^2+y^2+z^2)-(x^3y^2+x^3z^2+y^3x^2+y^3z^2+z^3x^2+z^3y^2)]$

$=2[3xyz(x^2+y^2+z^2)-x^2y^2(x+y)-y^2z^2(y+z)-z^2x^2(z+x)]$

$=6xyz(x^2+y^2+z^2)-2[x^2y^2(-z)+y^2z^2(-x)+z^2x^2(-y)]$

$=6xyz(x^2+y^2+z^2)+2(x^2y^2z+y^2z^2x+x^2x^2y)$

$=6xyz(x^2+y^2+z^2)+2xyz(xy+yz+xz)$

$=6xyz(x^2+y^2+z^2)+xyz[(x+y+z)^2-(x^2+y^2+z^2)]$

$=6xyz(x^2+y^2+z^2)+xyz[0-(x^2+y^2+z^2)]$

$=6xyz(x^2+y^2+z^2)-xyz(x^2+y^2+z^2)=5xyz(x^2+y^2+z^2)$

Ta có đpcm.

Từ giả thiết: \(x+y+z=0\Rightarrow x+y=-z\)

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

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=-c^3\)

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

\(\Leftrightarrow x^3+y^3+z^3=3xyz\) (1)

Nhận cả 2 vế của (1) với \(x^2+y^2+z^2\) ta được:

\(3xyz\left(x^2+y^2+z^2\right)=\left(x^2+y^2+z^2\right)\left(x^3+y^3+z^3\right)=x^5+x^3\left(y^2+z^2\right)+y^5+y^3\left(x^2+z^2\right)+z^5+z^3\left(x^2+y^2\right)\left(2\right)\)Do x + y + z =0 \(\Rightarrow y+z=-x\Rightarrow\left(y+z\right)^2=x^2\Leftrightarrow y^2+z^2=x^2-2yz\)Tương tự ta có:

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

Thay vào (2) ta được:

\(3xyz\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3\left(x^2-2yz\right)+y^3\left(y^2-2xz\right)+z^3\left(z^2-2xy\right)\)\(=2\left(x^5+y^5+z^5\right)-2xyz\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\left(đpcm\right)\)

Ta có: \(x+y+z=0\Leftrightarrow x+y=-z\)

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

\(\Leftrightarrow x^3-3xyz+y^3=-z^3\Leftrightarrow x^3+y^3+z^3=3xyz\)

Do đó \(3xyz\left(x^2+y^2+z^2\right)=\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)\)

\(=x^5+y^5+z^5+x^3\left(y^2+z^2\right)+y^3\left(z^2+x^2\right)+z^3\left(x^2+y^2\right)\) (*)

Mà \(x^2+y^2=\left(x+y\right)^2-2xy=\left(-z\right)^2-2xy=z^2-2xy\) (vì x + y = -z) (1)

Tương tự, ta có: \(y^2+z^2=x^2-2yz\left(2\right);z^2+x^2=y^2-2zx\left(3\right)\)

Thay (1);(2);(3) vào (*) ta được:

\(3xyz\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3\left(x^2-2yz\right)+y^3\left(y^2-2zx\right)+z^3\left(z^2-2xy\right)\)

\(=x^5+y^5+z^5+x^5-2x^3yz+y^5-2xy^3z+z^5-2xyz^3\)

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

\(\Leftrightarrow3xyz\left(x^2+y^2+z^2\right)+2xyz\left(x^2+y^2+z^2\right)=2\left(x^5+y^5+z^2\right)-2xyz\left(x^2+y^2+z^2\right)+2xyz\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow5xyz\left(x^2+y^2+z^2\right)=2\left(x^5+y^5+z^5\right)\left(đpcm\right)\)

tích mình đi

ai tích mình

mình ko tích lại đâu

thanks

Xét các biểu thức :

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

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

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

Do đó VT có giá trị là \(5.\left(3xyz\right).2\left(x^2+y^2+xy\right)=30xyz\left(x^2+y^2+xy\right)\)

Xét VP:

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

\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\right)\)

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

\(=-5xy.\left[\left(x+y\right)^3-xy\left(x+y\right)\right]\)

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

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

Do đó VP là \(30xyz\left(x^2+y^2+xy\right)\)

Suy ra điều phải chứng minh.

Ta có: x+y+z=0 => x³+y³+z³=3xyz (tự c/m)

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

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

Tương tự ta cũng có: \(y^2+z^2=x^2-2yz;z^2+x^2=y^2-2zx\)

Lại có: \(\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)\)

\(=x^5+x^3y^2+x^3z^2+y^3x^2+y^5+y^3z^2+z^3x^2+z^3y^2+z^5\)

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

\(=x^5+y^5+z^5+x^3\left(x^2-2yz\right)+y^3\left(y^2-2xz\right)+z^3\left(z^2-2xy\right)\)

\(=x^5+y^5+z^5+x^5-2x^3yz+y^5-2xy^3z+z^5-2xyz^3\)

\(\Rightarrow3xyz\left(x^2+y^2+z^2\right)=2\left(x^5+y^5+z^5\right)-2xyz\left(x^2+y^2+z^2\right)\)

\(\Rightarrow5xyz\left(x^2+y^2+z^2\right)=2\left(x^5+y^5+z^5\right)\) (đpcm)

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