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b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)
bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
Ta có:
x2+y2+z2=12;x+y+z=6⇒3(x2+y2+z2)−(x+y+z)2=0⇔3(x2+y2+z2)−(x2+y2+z2+2xy+2xz+2yz)=0⇔2x2+2y2+2z2−2xy−2xz−2yz=0⇔(x−y)2+(y−z)2+(z−x)2=0(1)x2+y2+z2=12;x+y+z=6⇒3(x2+y2+z2)−(x+y+z)2=0⇔3(x2+y2+z2)−(x2+y2+z2+2xy+2xz+2yz)=0⇔2x2+2y2+2z2−2xy−2xz−2yz=0⇔(x−y)2+(y−z)2+(z−x)2=0(1)
Mà (x−y)2,(y−z)2,(z−x)2≥0,∀x,y,z(x−y)2,(y−z)2,(z−x)2≥0,∀x,y,z
⇒(x−y)2+(y−z)2+(z−x)2≥0,∀x,y,z⇒(x−y)2+(y−z)2+(z−x)2≥0,∀x,y,z
→(1)→(1) đúng chỉ khi dấu bằng xảy ra
(x−y)2=(y−z)2=(z−x)2=0⇔x−y=y−z=z−x=0⇔x=y=z(x−y)2=(y−z)2=(z−x)2=0⇔x−y=y−z=z−x=0⇔x=y=z
Mà x+y+z=6x+y+z=6⇒x=y=z=2⇒x=y=z=2\, suy ra A=10
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.
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)
