cho x,y,z >0 thỏa mãn \(x^3+y^3+z^3=3xyz\)
tính \(P=\frac{\left(x-y\right)^{2017}}{x+y}+\frac{\left(y-z\right)^{2018}}{y+z}+\frac{\left(z-x\right)^{2019}}{x+z}\)
1/Cho a,b,c thỏa mãn \(\frac{2}{\left(x^2+1\right)\left(x-1\right)}=\frac{ax+b}{x^2+1}+\frac{c}{x-1}\)
Tính giá trị biểu thức M=\(\frac{a^{2017}+b^{2018}+c^{2019}}{a^{2017}b^{2018}c^{2019}}\)
2/Cho x,y,z≠0 và x+y+z=2008
Tính giá trị biểu thức P=\(\frac{x^3}{\left(x-y\right)\left(x-z\right)}+\frac{y^3}{\left(y-x\right)\left(y-z\right)}+\frac{z^3}{\left(z-y\right)\left(z-x\right)}\)
1/Cho a,b,c thỏa mãn \(\frac{2}{\left(x^2+1\right)\left(x-1\right)}=\frac{ax+b}{x^2+1}+\frac{c}{x-1}\)
Tính giá trị biểu thức M=\(\frac{a^{2017}+b^{2018}+c^{2019}}{a^{2017}b^{2018}c^{2019}}\)
2/Cho x,y,z≠0 và x+y+z=2008
Tính giá trị biểu thức P=\(\frac{x^3}{\left(x-y\right)\left(x-z\right)}+\frac{y^3}{\left(y-x\right)\left(y-z\right)}+\frac{z^3}{\left(z-y\right)\left(z-x\right)}\)
Cho 3 số x,y,z khác 0 đồng thời thỏa mãn \(x+y+z=\frac{1}{2},\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xyz}=4\) và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\)
Tính giá trị biểu thức Q=\(\left(y^{2017}+z^{2017}\right)\left(z^{2019}+x^{2019}\right)\left(x^{2021}+y^{2021}\right)\)
Cho x,y,z là 3 số khác 0 thỏa mãn điều kiện x3+y3+z3=3xyz và x+y+z=0.Tính giá trị của biểu thức:
\(M=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
X3 + Y3 + Z3 = 3XYZ
<=> X3 + Y3 + Z3 - 3XYZ = 0
<=> ( X3 + Y3 ) + Z3 - 3XYZ = 0
<=> ( X + Y )3 - 3XY( X + Y ) + Z3 - 3XYZ = 0
<=> [ ( X + Y )3 + Z3 ] - 3XY( X + Y + Z ) = 0
<=> ( X + Y + Z )[ ( X + Y )2 - ( X + Y ).Z + Z2 - 3XY ] = 0
<=> ( X + Y + Z )( X2 + Y2 + Z2 - XY - YZ - XZ ) = 0
<=> \(\orbr{\begin{cases}X+Y+Z=0\\X^2+Y^2+Z^2-XY-YZ-XZ=0\end{cases}}\)
+) X + Y + Z = 0 => \(\hept{\begin{cases}X+Y=-Z\\Y+Z=-X\\X+Z=-Y\end{cases}}\)
KHI ĐÓ : \(M=\left(1+\frac{X}{Y}\right)\left(1+\frac{Y}{Z}\right)\left(1+\frac{Z}{X}\right)=\left(\frac{X+Y}{Y}\right)\left(\frac{Y+Z}{Z}\right)\left(\frac{X+Z}{X}\right)=\frac{-Z}{Y}\cdot\frac{-X}{Z}\cdot\frac{-Y}{X}=-1\)
+) X2 + Y2 + Z2 - XY - YZ - XZ = 0
<=> 2( X2 + Y2 + Z2 - XY - YZ - XZ ) = 0
<=> 2X2 + 2Y2 + 2Z2 - 2XY - 2YZ - 2XZ = 0
<=> ( X2 - 2XY + Y2 ) + ( Y2 - 2YZ + Z2 ) + ( X2 - 2XZ + Z2 ) = 0
<=> ( X - Y )2 + ( Y - Z )2 + ( X - Z )2 = 0 (1)
DỄ DÀNG CHỨNG MINH (1) ≥ 0 ∀ X,Y,Z
DẤU "=" XẢY RA <=> X = Y = Z
KHI ĐÓ : \(M=\left(1+\frac{X}{Y}\right)\left(1+\frac{Y}{Z}\right)\left(1+\frac{Z}{X}\right)=\left(1+\frac{Y}{Y}\right)\left(1+\frac{Z}{Z}\right)\left(1+\frac{X}{X}\right)=2\cdot2\cdot2=8\)
Khi x + y + z = 0
=> x + y = -z
=> x + z = - y
=> y + z = - x
Khi đó M = \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}=-1\)
cho x,y,z thỏa mãn \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\left(\frac{1}{x+y+z}\right)=1\)
tìm B=\(\left(x^{2016}+y^{2016}\right)\left(y^{2017}+z^{2017}\right)\left(z^{2018}+x^{2018}\right)\)
Bài 1:Cho x,y,z là 3 số khác 0.thỏa mãn \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)\(1\)
TÍNH GT BT
\(A=\left(x^{25}+y^{25}\right)\left(y^3+z^3\right)\left(x^{2019}+z^{2019}\right)\)
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\Rightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)\(\Rightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)\(\Rightarrow\)\(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)
\(\Rightarrow A=0\)
Sai đề thật, cái biểu thức trên 100% lớn hơn hoặc = 9, lấy đâu ra =1
Cho x,y,x là 3 số thực khác 0 thỏa mãn \(\hept{\begin{cases}x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{x}+\frac{1}{z}\right)+z\left(\frac{1}{y}+\frac{1}{x}\right)=-2\\x^3+y^3+z^3=1\end{cases}}\)
Tính \(P=\frac{1}{x^{2017}}+\frac{1}{y^{2017}}+\frac{1}{z^{2017}}\)
cho các số thỏa mãn x+y+z=1000
tính \(Q=\frac{x^3}{\left(x-y\right).\left(x-z\right)}+\frac{y^3}{\left(y-x\right)\left(y-z\right)}+\frac{z^3}{\left(z-x\right)\left(z-y\right)}\)
\(Q=\frac{x^3}{\left(x-y\right)\left(x-z\right)}+\frac{y^3}{\left(y-x\right)\left(y-z\right)}+\frac{z^3}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^3}{\left(x-y\right)\left(x-z\right)}-\frac{y^3}{\left(x-y\right)\left(y-z\right)}+\frac{z^3}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^3\left(y-z\right)-y^3\left(x-z\right)+z^3\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)(1)
Ta có:
\(x^3\left(y-z\right)-y^3\left(x-z\right)+z^3\left(x-y\right)\)
\(=x^3\left(y-z\right)-y^3\left(y-z\right)-y^3\left(x-y\right)+z^3\left(x-y\right)\)
\(=\left(y-z\right)\left(x^3-y^3\right)-\left(x-y\right)\left(y^3-z^3\right)\)
\(=\left(y-z\right)\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x-y\right)\left(y-z\right)\left(y^2+yz+z^2\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x^2+xy+y^2-y^2-yz-z^2\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x^2+xy-yz-z^2\right)\)
\(=\left(x-y\right)\left(y-z\right)\left[\left(x-z\right)\left(x+z\right)+y\left(x-z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\left(x+y+z\right)=1000\left(x-y\right)\left(y-z\right)\left(x-z\right)\)(2)
Từ (1) và (2), ta có Q = 1000
Cho x, y, z đôi một khác nhau thỏa mãn: \(x^3+y^3+z^3=3xyz\) và \(xyz\ne0\). Tính: \(B=\dfrac{16.\left(x+y\right)}{z}+\dfrac{3.\left(y+z\right)}{x}-\dfrac{2019.\left(x+z\right)}{y}\)
\(x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)