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20 tháng 3 2021
Ta có: \(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)^3-3.\left(x+y\right).z.\left(x+y+z\right)-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right).\left[\left(x+y+z\right)^2-3.\left(x+y\right).z\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2zx-3xz-3yz-3xy\right)=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(x^2+y^2+z^2-xz-yz-xy\right)=0\)
+ \(x+y+z=0\)\(\Rightarrow\)\(C=\frac{x^{2019}+y^{2019}+z^{2019}}{0}\)( Loại )
+ \(x^2+y^2+z^2-xz-yz-xy=0\)
\(\Rightarrow2x^2+2y^2+2z^2-2xz-2yz-2xy=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Rightarrow\)\(x=y=z\)
\(\Rightarrow\)\(C=\frac{x^{2019}+x^{2019}+x^{2019}}{\left(x+x+x\right)^{2019}}=\frac{3.x^{2019}}{3^{2019}.x^{2019}}=\frac{1}{3^{2018}}\)
Vậy.......
20 tháng 3 2021
Từ x3 + y3 + z3 = 3xyz
=> ( x + y + z )( x2 + y2 + z2 - xy - yz - xz ) = 0 ( phân tích như bạn kia )
Vì x + y + z ≠ 0
=> x2 + y2 + z2 - xy - yz - xz = 0
<=> 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz = 0
<=> ( x - y )2 + ( y - z )2 + ( x - z )2 = 0
VT ≥ 0 ∀ x,y,z. Đẳng thức xảy ra <=> x=y=z
Khi đó \(C=\frac{x^{2019}+y^{2019}+z^{2019}}{\left(x+y+z\right)^{2019}}=\frac{3x^{2019}}{\left(3x\right)^{2019}}=\frac{3x^{2019}}{3^{2019}\cdot x^{2019}}=\frac{1}{3^{2018}}\)
PA
17 tháng 11 2016
\(x^3+y^3+z^3=3xyz\)
\(x^3+y^3+z^3-3xyz=0\)
\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
\(x^2+y^2+z^2-xy-xz-yz=0\left(x+y+z\ne0\right)\)
\(2\times\left(x^2+y^2+z^2-xy-xz-yz\right)=0\times2\)
\(2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)
\(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)
\(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)
\(\left[\begin{array}{nghiempt}x-y=0\\x-z=0\\y-z=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=y\\x=z\\y=z\end{array}\right.\)
x = y = z
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{x}{z}\right)\)
\(=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)
\(=2^3\)
\(=8\)
2 tháng 6 2022
Bài 1:
a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)
Để A=0 thì x+1=0
hay x=-1
b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)
Để B=0 thi (x-2)(x+2)=0
=>x=2 hoặc x=-2
FA
6 tháng 10 2017
\(A=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x^2+2xy+y^2\right)-\left(xz+yz\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(=0\)
<><><>
\(A=\left(\dfrac{x}{y}+1\right)\left(\dfrac{y}{z}+1\right)\left(\dfrac{z}{x}+1\right)\)
\(=\dfrac{x+y}{y}\times\dfrac{y+z}{z}\times\dfrac{z+x}{x}\)
\(=\dfrac{-z}{y}\times\dfrac{-x}{z}\times\dfrac{-y}{x}\)
\(=-1\)
<><><>
\(A=\dfrac{1}{y^2+z^2-x^2}+\dfrac{1}{x^2+z^2-y^2}+\dfrac{1}{x^2+y^2-z^2}\)
\(=\dfrac{1}{\left(y+z\right)^2-2yz-x^2}+\dfrac{1}{\left(x+z\right)^2-2xz-y^2}+\dfrac{1}{\left(x+y\right)^2-2xy-z^2}\)
\(=\dfrac{1}{\left(-x\right)^2-2yz-x^2}+\dfrac{1}{\left(-y\right)^2-2xz-y^2}+\dfrac{1}{\left(-z\right)^2-2xy-z^2}\)
\(=-\dfrac{1}{2}\left(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xz}\right)\)
\(=-\dfrac{1}{2}\times\dfrac{x+y+z}{xyz}\)
\(=0\)
16 tháng 11 2016
Ta có
x3 + y3 + z3 - 3xyz = 0
<=> (x + y)3 + z3 - 3xy(x + y) - 3xyz = 0
<=> (x + y + z)(x2 + y2 + z2 - xy - yz - xz) = 0
Mà theo đề bài (x + y + z) \(\ne\)0 nên
(x2 + y2 + z2 - xy - yz - xz) = 0
Ta có x2 + y2 + z2 - xy - yz - xz \(\ge\)xy + yz + xz - xy - yz - xz = 0
Dấu = xảy ra khi x = y = z
Từ đó ta có
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
9 tháng 8 2016
x^3 + y^3 + z^3 - 3xyz = (x+y)^3 + z^3 - 3x^2y - 3xy^2 - 3xyz
= (x+y)^3 + z^3 - 3xy(x + y + z)
= (x+y+z)^3 - 3(x+y)^2.z - 3(x+y)z^2 - 3xy(x + y + z)
= (x+y+z)^3 - 3(x+y)z(x+ y + z) - 3xy(x + y + z)
=(x+y+z)[(x+y+z)^2 - 3(x+y)z - 3xy]
=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)
=1/2(x+y+z)(x^2-2xy+y^2+y^2-2yz+z^2+x^2-2xz+z^2)
=1/2(x+y+z)[(x-y)^2+(y-z)^2+(x-z)^2]
mà x^3 + y^3 + z^3 - 3xyz=0
<=> x+y+z=0
Vậy ...
Chúc bạn học tốt .
hoặc (x-y)^2+(y-z)^2+(x-z)^2 =0 mà (x-y)^2,(y-z)^2,(x-z)^2 >=0 mọi x,y,z
=> x-y=y-z=x-z=0 => x=y=z
17 tháng 10 2020
a) Ta có: \(27x^3+\frac{y^3}{8}\)
\(=\left(3x\right)^3+\left(\frac{y}{2}\right)^3\)
\(=\left(3x+\frac{y}{2}\right)\left(9x^2-\frac{3xy}{2}+\frac{y^2}{4}\right)\)
b) Ta có: \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
c) Ta có: \(x^{m+2}+x^m\)
\(=x^m\cdot x^2+x^m\)
\(=x^m\left(x^2+1\right)\)
d) Ta có: \(x^{k+1}-x^{k-1}\)
\(=x^{k-1}\cdot x^2-x^{k-1}\cdot1\)
\(=x^{k-1}\left(x^2-1\right)\)
\(=x^{k-1}\cdot\left(x-1\right)\left(x+1\right)\)
f) Ta có: \(\left(a+b-c\right)\cdot x^2-\left(c-a-b\right)x\)
\(=x^2\left(a+b-c\right)+x\left(a+b-c\right)\)
\(=x\left(a+b-c\right)\left(x+1\right)\)
e) Ta có: \(\left(a-2b\right)^{3n+1}\)
\(=\left(a-2b\right)^{3n}\cdot\left(a-2b\right)\)
n) Ta có: \(\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x^3+y^3\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)
\(=3xy\left(x+y\right)\)
Sửa đề: Chứng minh x=y=z
\(x^3+y^3+z^3=3xyz\)
=>\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
=>\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
=>\(\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz-3xy\right)=0\)
=>\(x^2+y^2+z^2-xy-xz-yz=0\)
=>\(2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)
=>\(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)
=>\(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
=>x=y=z
Có:
\(x^3+y^3+z^3=3xyz\\\Leftrightarrow x^3+y^3+z^3-3xyz=0\\\Leftrightarrow(x+y)^3+z^3-3xy(x+y)-3xyz=0\\\Leftrightarrow (x+y+z)^3-3(x+y)z(x+y+z)-3xy(x+y+z)=0\\\Leftrightarrow (x+y+z)[(x+y+z)^2-3(x+y)z-3xy]=0\\\Leftrightarrow (x+y+z)(x^2+y^2+z^2+2xy+2yz+2xz-3xz-3yz-3xy)=0\\\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0\\\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0 (vì.x+y+z\neq0)\\\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\\\Leftrightarrow(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)=0\\\Leftrightarrow(x-y)^2+(y-z)^2+(x-z)^2=0\)
Ta thấy: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x;y\\\left(y-z\right)^2\ge0\forall x;y\\\left(x-z\right)^2\ge0\forall x;y\end{matrix}\right.\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x;y;z\)
Mà: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
nên: \(\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\x=z\end{matrix}\right.\Leftrightarrow x=y=z\left(đpcm\right)\)
\(Toru\)