Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(x+y)^3 - 3xy(x+y) + z^3 - 3xyz = 0
(x+y+z) ( (x+y)^2 +z^2 -z(x+y) -3xy) =0
(x+y+z) ( x^2+ 2xy+y^2 +z^2- zx-zy-3xy)=0
(x+y+z) ( x^2+y^2+z^2 -zx-zy -xy)=0
Suy ra x+y+z =0
x+y = -z
y+z = -x
x+z = -y
B = -16 + (-3) +2038 = 2019
Ta có: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\left(x,y,z\ne0\right)\)
+) x + y + z = 0 \(\Rightarrow B=\frac{-16z}{z}+\frac{-3x}{x}-\frac{-2038y}{y}\)
\(=-16-3+2038=2019\)
+) x = y = z \(\Rightarrow B=\frac{16.2z}{z}+\frac{3.2x}{x}-\frac{2038.2y}{y}\)
\(=32+6-4076=-4038\)
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.......
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 \(C=\frac{x^{2019}+y^{2019}+z^{2019}}{\left(x+y+z\right)^{2019}}=\frac{3x^{2019}}{\left(3x\right)^{2019}}=\frac{3x^{2019}}{3^{2019}\cdot x^{2019}}=\frac{1}{3^{2018}}\)
Ta có:
\(x^3+y^3+z^3=3xyz\left(gt\right)\)
\(\Rightarrow x^3+y^3+z^3-3xyz=0\)
\(\Rightarrow x^3+y^3+3xy\left(x+y\right)+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Rightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Rightarrow\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x+y+z\right)=0\)
\(\Rightarrow\left(x+y+z\right)^3-\left(x+y+z\right)\left(3xy+3zx+3yz\right)=0\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xy-3xz-3yz\right)=0\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
\(\Rightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y+z=0\\\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\end{matrix}\right.\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
Mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)
\(\Rightarrow x=y=z\)
Xét trường hợp x = y = z, ta có:
\(P=\dfrac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(P=\dfrac{x^3}{2x.2x.2x}\)
\(P=\dfrac{x^3}{8x^3}\)
\(P=\dfrac{1}{8}\)
Xét trường hợp x + y + z = 0, ta có:
\(\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(y+x\right)\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{-\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\Rightarrow P=-1\)
phân tích gt sau đó suy ra x+y+x=0
từ đây tính đc x+y=? y+z=? x+z=?
ta được kết quả là'; -2006
Xét \(x^3+y^3+z^3=3xyz\)
\(x^3+y^3+z^3-3xyz=0\)
\(\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)
\(\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\left(x+y+z\right)\left(x^2+2xy+y^2-xy-yz+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
TH1:\(x+y+z=0\)
\(\Rightarrow x+y=-z;y+z=-x;z+x=-y\left(1\right)\)
Thay (1) vô pt cần tính:
\(\frac{2016xyz}{-z.-x.-y}=\frac{2016xyz}{-\left(xyz\right)}=-2016\)
TH2:\(x^2+y^2+z^2-xy-yz-xz=0\)
Nhân 2 vế với 2
\(2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)
\(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)
Do VT dương
\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-z\right)^2=0\\\left(y-z\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x-y=0\\x-z=0\\y-z=0\end{cases}\Rightarrow}\hept{\begin{cases}x=y\\x=z\\y=z\end{cases}}\Rightarrow x=y=z\)
Thay y,z ở pt cần tính là x
\(\Rightarrow\frac{2016x.x.x}{\left(x+x\right)\left(x+x\right)\left(x+x\right)}=\frac{2016x^3}{2x.2x.2x}=\frac{2016x^3}{8x^3}=\frac{2016}{8}=252\)
Vậy pt có thể = -2016 khi x + y + z = 0
pt có thể = 252 khi \(x^2+y^2+z^2-xy-xz-yz=0\)
\(A=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x^2+2xy+y^2\right)-\left(xz+yz\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(=0\)
<><><>
\(A=\left(\dfrac{x}{y}+1\right)\left(\dfrac{y}{z}+1\right)\left(\dfrac{z}{x}+1\right)\)
\(=\dfrac{x+y}{y}\times\dfrac{y+z}{z}\times\dfrac{z+x}{x}\)
\(=\dfrac{-z}{y}\times\dfrac{-x}{z}\times\dfrac{-y}{x}\)
\(=-1\)
<><><>
\(A=\dfrac{1}{y^2+z^2-x^2}+\dfrac{1}{x^2+z^2-y^2}+\dfrac{1}{x^2+y^2-z^2}\)
\(=\dfrac{1}{\left(y+z\right)^2-2yz-x^2}+\dfrac{1}{\left(x+z\right)^2-2xz-y^2}+\dfrac{1}{\left(x+y\right)^2-2xy-z^2}\)
\(=\dfrac{1}{\left(-x\right)^2-2yz-x^2}+\dfrac{1}{\left(-y\right)^2-2xz-y^2}+\dfrac{1}{\left(-z\right)^2-2xy-z^2}\)
\(=-\dfrac{1}{2}\left(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xz}\right)\)
\(=-\dfrac{1}{2}\times\dfrac{x+y+z}{xyz}\)
\(=0\)
dat a=x-y
b=y-z
c=z-x
a+b+c=0=x+y+z
\(\left(\frac{a}{z}+\frac{b}{x}+\frac{c}{y}\right)\left(\frac{z}{a}+\frac{x}{b}+\frac{y}{c}\right)\)
dung bumiakopsky de giai
...........................................
1: \(=\dfrac{\left(x^2+2xy+y^2\right)-1}{\left(x^2+2x+1\right)-y^2}\)
\(=\dfrac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}=\dfrac{x+y-1}{x-y+1}\)
2: \(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{x^2-xy+y^2}\)
3: \(=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{x+y+z}{2}\)
\(x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)
GIÁO VIÊN SAO TOÀN SAI HẰNG ĐẲNG THỨC THẾ????