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a)Đặt A=(x+y+z)3-x3-y3-z3
Xét (x+y+z)3=[(x+y)+z]3=(x+y)3+z3+3z(x+y)(x+y+z) =x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)
=(x3+y3+z3)+3(x+y)(xy+xz+yz+z2)
=(x3+y3+z3)+3(x+y)[(xy+yz)+(xz+z2)]
=(x3+y3+z3)+3(x+y)[y(x+z)+z(x+z)]
=(x3+y3+z3)+3(x+y)(x+z)(y+z)
Từ đó suy ra A=(x3+y3+z3)+3(x+y)(x+z)(y+z)-x3-y3-z3=3(x+y)(x+z)(y+z)
x^3 + y^3 + z^3 +3(x+y)(y+z)(z+x)=x3+y3+z3+(3x+3y)(y+z)(z+x)
=x3+y3+z3+(3xy+3xz+3y2+3yz)(z+x)
=x3+y3+z3+3xyz+3x2y+3xz2+3x2z+3y2z+3y2x+3yz2+3xyz
=x3+y3+z3+3x2y+3xz2+3x2z+3y2z+3y2x+3yz2+6xyz
=x3+3x2y+3y2x+y3+3x2z+6xyz+3y2z+3xz2+3yz2+z3
=(x+y)3+3z(x2+2xy+y2)+3z2(x+y)+z3
=(x+y)3+3z(x+y)2+3z2(x+y)+z3
=(x+y+z)3
vậy (x+y+z)^3= x^3 + y^3 + z^3 +3(x+y)(y+z)(z+x)
a, \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Leftrightarrow\left(x+y\right)^3=-z^3\)
\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=-z^3\)
\(\Leftrightarrow x^3+y^3+z^3=-3xy\left(x+y\right)\)
\(\Leftrightarrow x^3+y^3+z^3=3xyz\)(vì x+y=-z)
Bài 3:
\(\left\{{}\begin{matrix}x+y>=2\sqrt{xy}\\y+z>=2\sqrt{yz}\\x+z>=2\sqrt{xz}\end{matrix}\right.\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)>=8xyz\)
Dấu = xảy ra khi x=y=z
\(x+y+z=0\Rightarrow z=-\left(x+y\right)\)
\(x^3+x^2z+y^2z-xyz+y^3=x^3+y^3+\left(x^2+y^2-xy\right)z\)
\(=x^3+y^3-\left(x+y\right)\left(x^2+y^2-xy\right)\)
\(=x^3+y^3-\left(x^3+y^3\right)=0\)
x+y+z=0⇒z=−(x+y)�+�+�=0⇒�=−(�+�)
x3+x2z+y2z−xyz+y3=x3+y3+(x2+y2−xy)z�3+�2�+�2�−���+�3=�3+�3+(�2+�2−��)�
=x3+y3−(x+y)(x2+y2−xy)=�3+�3−(�+�)(�2+�2−��)
=x3+y3−(x3+y3)=0
Đặt A=(x+y+z)3-x3-y3-z3
Xét (x+y+z)3=[(x+y)+z]3=(x+y)3+z3+3z(x+y)(x+y+z)=x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)
=(x3+y3+z3)+3(x+y)(xy+xz+yz+z2)
=(x3+y3+z3)+3(x+y)[(xy+yz)+(xz+z2)]
=(x3+y3+z3)+3(x+y)[y(x+z)+z(x+z)]
=(x3+y3+z3)+3(x+y)(x+z)(y+z)
Từ đó suy ra A=(x3+y3+z3)+3(x+y)(x+z)(y+z)-x3-y3-z3=3(x+y)(x+z)(y+z
\(VT=\left(x+y+z\right)^3=\left[\left(x+y\right)+z\right]^3\)
\(=\left(x+y\right)^3+z^3+3\left(x+y\right)z\left(x+y+z\right)\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3\left(x+y\right)z\left(x+y+z\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
\(=VP\left(đpcm\right)\)
\(\left(x+y+z\right)^3=x^3+y^3+z^3+3x^2y+3xy^2+3y^2z+3z^2x+3x^2z+3z^2x+6xyz\)
=\(x^3+y^3+z^3+3\left(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+2xyz\right)\)
=\(x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)(đpcm)