cho \(x+y+z=0\)cm rằng:\(2.\left(x^5+y^5+z^5\right)=5xyz.\left(x^2+y^2+z^2\right)\)
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NM
1
AH
Akai Haruma
Giáo viên
27 tháng 8 2023
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.
NN
1
TN
2 tháng 7 2017
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:
\(3xyz\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3\left(x^2-2yz\right)+y^3\left(y^2-2xz\right)+z^3\left(z^2-2xy\right)\)\(=2\left(x^5+y^5+z^5\right)-2xyz\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\left(đpcm\right)\)
ND
7
S
7 tháng 7 2018
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:
\(3xyz\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3\left(x^2-2yz\right)+y^3\left(y^2-2zx\right)+z^3\left(z^2-2xy\right)\)
\(=x^5+y^5+z^5+x^5-2x^3yz+y^5-2xy^3z+z^5-2xyz^3\)
\(=2\left(x^5+y^5+z^5\right)-2xyz\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow3xyz\left(x^2+y^2+z^2\right)+2xyz\left(x^2+y^2+z^2\right)=2\left(x^5+y^5+z^2\right)-2xyz\left(x^2+y^2+z^2\right)+2xyz\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow5xyz\left(x^2+y^2+z^2\right)=2\left(x^5+y^5+z^5\right)\left(đpcm\right)\)
29 tháng 3 2020
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)
NQ
2 tháng 4 2020
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)
31 tháng 3 2020
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J
1
4 tháng 7 2017
\(y+z=-x\)
\(\left(y+z\right)^5=-x^5\)
\(y^5+5y^4z+10y^3z^2+10y^2z^3+5yz^4+z^5+x^5=0\)
\(x^5+y^5+z^5+5yz\left(y^3+2y^2z+2yz^2+z^3\right)=0\)
\(x^5+y^5+z^5+5yz\left[\left(y+z\right)\left(y^2-yz+z^2\right)+2yz\left(y+z\right)\right]=0\)
\(x^5+y^5+z^5+5yz\left(y+z\right)\left(y^2+yz+z^2\right)=0\)
\(2\left(x^5+y^5+z^5\right)-5xyz\left(\left(y^2+2yz+z^2\right)+y^2+z^2\right)=0\)
\(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
TC
0
TH
6
A
0
vì \(x+y+z=0\) \(\Rightarrow\)\(x+y=-z\)
\(\Rightarrow\left(x+y\right)^3=-z^3\)\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)
\(\Rightarrow x^3+y^3+z^3=-3xy.\left(x+y\right)\)
\(\Rightarrow x^3+y^3+z^3=3xyz\) do \(x+y=-z\)
\(\Rightarrow\left(x^3+y^3+z^3\right).\left(x^2+y^2+z^2\right)=3xyz.\left(x^2+y^2+z^2\right)\)
\(\Rightarrow3xyz.\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3.\left(y^2+z^2\right)+y^3.\left(x^2+z^2\right)+z^3.\left(x^2+y^2\right)\)
lại có: \(x^2+y^2=\left(x+y\right)^2-2xy=z^2-2xy\)
tương tự thì: \(y^2+z^2=x^2-2yz\)
\(z^2+x^2=y^2-2xz\)
vì vậy nên \(3xyz.\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3.\left(x^2-2yz\right)+y^3.\left(y^2-2xz\right)+z^3.\left(z^2-2xy\right)\)
\(\Rightarrow3xyz.\left(x^2+y^2+z^2\right)=2x^5+2y^5+2z^5-2xyz.\left(x^2+y^2+z^2\right)\)
\(\Rightarrow5xyz.\left(x^2+y^2+z^2\right)=2.\left(x^5+y^5+z^5\right)\)
đpcm
VÌ \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow\left(x+y\right)^5=-z^5\)
\(\Leftrightarrow x^5+y^{^5}+5\left(x^4y+xy^4+2x^3y^2+2x^2y^3\right)=-z^5\)
\(\Leftrightarrow x^5+y^{^5}+z^5+5xy\left(x^3+y^3+2x^3y^2+2x^2y^3\right)=0\)
\(\Leftrightarrow x^5+y^{^5}+z^5+5xy\left(x+y\right)+\left(x^2-xy+y^2+2xy\right)=0\)
\(\Leftrightarrow x^5+y^{^5}+z^5-5xyz\left(x^2+xy+y^2\right)=0\)
\(\Leftrightarrow x^5+y^{^5}+z^5=5xyz\left(x^2+xy+y^2\right)\)
\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(2x^2+2xy+2y^2\right)\)
\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+\left(x+y\right)^2+y^2\right)\)
\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)Vì (x+y=-z)
HT