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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 \(x^2+y^2+z^2-xy-xz-yz=0\)
Đề chưa chuẩn: tuy nhiên đánh vào -2016 => đáp án đúng:
Vì bản chất như sau:
thỏa ĐK ban đầu x^3+y^3+z^3=3xzy
Từ HĐT=>
\(\orbr{\begin{cases}x+y+z=0\left(1\right)\\x^2+y^2+z^2-xy-yz-xz=0\left(2\right)\end{cases}}\)
=>(1)&(2) đều có cặp nghiệm x=y=z=0 khi đó P không xác định
do vậy đề thiếu điều kiện x,y,z không đồng thời =0:(*)
Nếu thêm đk (*) giải tiếp
(2) vô nghiệm
do vậy khi đó chỉ có nghiệm duy nhất của (1)
x+y=-z
x+z=-y
z+y=-x
Thay vào biểu thwucs P=-2016
\(x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)
\(\dfrac{x^3+y^3+z^3-3xyz}{xy^2+xz\left(2y+z\right)}.\dfrac{x\left(x+y\right)+y\left(x-xy\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)}{xy^2+2xyz+x^2z}.\dfrac{x^2+xy-xy-xy^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\\ =\dfrac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2xy^2+4xyz+2x^2z}.\dfrac{x^2-xy^2}{\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2}\\ =\dfrac{\left(x+y+z\right)\left(x^2-xy\right)}{2xy^2+4xy+2x^2z}\)
@@ ko ra nữa
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:
\(\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(y+x\right)\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{-\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\Rightarrow P=-1\)
\(=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-xz-yz\right)}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)}{2\left(x^2+y^2+z^2-xy-xz-yz\right)}\)
\(=\dfrac{x+y+z}{2}\)
Ta có : \(B=\dfrac{\left(x+z\right)^3-y^3-3xz\left(x+z\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2+z^2+2xz+yz+xy+y^2\right)-3xz\left(x-y+z\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\dfrac{1}{2}\left(x-y+z\right)2\left(x^2+y^2+z^2+xy+zy-xz\right)}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\dfrac{1}{2}\left(x-y+z\right)\left[\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\right]}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{x-y+z}{2}\)
\(x^3+y^3+z^3=3xyz\)
⇔ \(\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz=0\)
⇔ \(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
⇔ \(\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)
⇔ \(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
⇔ \(\left(x+y+z\right)\left(x^2-2xy+y^2+z^2-2xz+x^2+y^2-2yz+z^2\right)=0\) ⇔ \(\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
Do : x , y , z là ba số thực phân biệt , ta có :
\(x+y+z=0\)
⇔ \(x+y=-z;y+z=-x;x+z=-y\)
Khi đó , ta có : \(P=\dfrac{2016xyz}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\dfrac{2016xyz}{-xyz}=-2016\)