CM:\(B=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2,\) la so chinh phuong \(\left(x,y,z\in Z\right)\)
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T
25 tháng 11 2019
\(A=\left(x^2+y^2+z^2\right)\left[\left(x^2+y^2+z^2\right)+2\left(xy+yz+zx\right)\right]+\left(xy+yz+zx\right)^2\)
\(=\left(x^2+y^2+z^2\right)^2+2\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)+\left(xy+yz+zx\right)^2\)
\(=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\) là một số chính phương (đpcm)
27 tháng 7 2017
b, \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left[\left(x-y\right)+\left(z-x\right)\right]+\left(z-x\right)^2\left(z-x\right)\)
\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left(x-y\right)-\left(y-z\right)^2\left(z-x\right)+\left(z-x\right)^2\left(z-x\right)\)
\(=\left(x-y\right)\left[\left(x-y\right)^2-\left(y-z\right)^2\right]-\left(z-x\right)\left[\left(y-z\right)^2-\left(z-x\right)^2\right]\)
\(=\left(x-y\right)\left(x-y-y+z\right)\left(x-y+y-z\right)-\left(z-x\right)\left(y-z-z+x\right)\left(y-z+z-x\right)\)
\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(z-x\right)\left(y-2z+x\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(x-z\right)\left(y-2z+x\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(x-z\right)\left(x-2y+z-y+2z-x\right)\)
\(=\left(x-y\right)\left(x-z\right)\left(3z-3y\right)\)
\(=3\left(x-y\right)\left(x-z\right)\left(z-y\right)\)
c, \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left[\left(y-x\right)-\left(z-x\right)\right]-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left(y-x\right)+y^2z^2\left(z-x\right)-z^2x^2\left(z-x\right)\)
\(=\left(x^2y^2-y^2z^2\right)\left(y-x\right)+\left(y^2z^2-z^2x^2\right)\left(z-x\right)\)
\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)+z^2\left(y-x\right)\left(x+y\right)\left(z-x\right)\)
\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)-z^2\left(y-x\right)\left(x+y\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(y-x\right)\left[y^2\left(x+z\right)-z^2\left(x+y\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left(y^2x+y^2z-z^2x-z^2y\right)\)
\(=\left(x-z\right)\left(y-x\right)\left[x\left(y^2-z^2\right)+yz\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left[x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left(y-z\right)\left(xy+xz+yz\right)\)
d, \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
7 tháng 7 2023
Phân tích vế trái ta được: 2(x2 + y2 + z2 − (xy + yz + zx)
Phân tích vế phải ta được: 6(x2 + y2 + z2 − (xy + yz + zx)
Vì VT = VP nên VP - VT=0
→ 4(x2 + y2 + z2 − (xy + yz + zx)) = 0
→2(2 (x2 + y2 + z2 − (xy + yz + zx))) = 0
→2((x − y)2 + (y − z)2 + (z − x)2) = 0
→(x − y)2 + (y − z)2 + (z − x)2 = 0
→(x − y)2 = 0; (y − z)2 = 0; (z − x)2 = 0
→x = y = z
S
23 tháng 12 2018
\(Taco:\)
\(\left(x+y\right)\left(y+z\right)=187\Leftrightarrow xy+xz+yy+yz=187\)
\(\left(y+z\right)\left(z+x\right)=154\Leftrightarrow yz+xy+zz+xz=154\)
\(\left(z+x\right)\left(x+y\right)=238\Leftrightarrow xz+zy+xx+xy=238\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)+\left(x+z\right)\left(x+y\right)+\left(y+z\right)\left(z+x\right)=579\)
\(\Leftrightarrow xy+zx+yy+yz+yz+xy+zz+xz+xz+zy+xx+xy=579\)
\(\Leftrightarrow3\left(xz+xy+yz\right)+x^2+y^2+z^2=579\)
\(\left(z+x\right)\left(x+y\right)-\left(x+y\right)\left(y+z\right)=51\)
\(\Leftrightarrow\left(x+y\right)\left(x-y\right)=x^2-y^2=51\)
\(\left(z+x\right)\left(x+y\right)-\left(y+z\right)\left(x+z\right)=84\)
\(\Leftrightarrow\left(x+z\right)\left(x-z\right)=84\Leftrightarrow x^2-z^2=84\)
\(\Leftrightarrow y^2-z^2=33\)
đến đây tịt
AH
Akai Haruma
Giáo viên
28 tháng 10 2021
Vì bài dài nên mình sẽ tách ra nhé.
1a. Ta có:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$
$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$
$=-3(-z)(-x)(-y)=3xyz$
$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$
------------------------
$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$
$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$
$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$
$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$
$=-z^5+5xyz^3-5x^2y^2z$
$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$
$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$
Từ $(1);(2)$ ta có đpcm.
AH
Akai Haruma
Giáo viên
28 tháng 10 2021
1b.
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$
$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$
Do đó:
$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$
$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$
$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$
$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$
$=7xyz(x^2y^2-2xyz^2+z^4)$
$=7xyz(xy-z^2)$
$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$
$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$
$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)
Ta có:
\(B=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2\)
\(=4\left(x^2+xy+xz\right)\left(x^2+xy+yz+xz\right)+y^2z^2\)
Đặt \(x^2+xy+xz=a\)
Khi đó: B trở thành:
\(4a\left(a+yz\right)+y^2z^2\)
\(=\left(yz+2a\right)^2\)
Hay \(B=\left(2x^2+2xy+2xz+yz\right)^2\)là số chính phương