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Ta có :x + y + z = -1 \(\Rightarrow\)x + y =-( 1 + z )
xy + yz + xz = 0 \(\Rightarrow\)xy = - z ( x + y ) = z ( z + 1 )
Tương tự : xz = y ( y + 1 ) ; yz = x . ( x + 1 )
\(M=\frac{z\left(z+1\right)}{z}+\frac{y\left(y+1\right)}{y}+\frac{x\left(x+1\right)}{x}=x+y+z+3=2\)
Bạn thay y xyz=2010 vào A ta được
A= xyz*x/xy+xyz*x+xyz + y/yz+y+xyz + z/xz+z+1
suy ra A=x^2yz/xy(1+xz+z) + y/y(z+1+xz) + z/xz+x+1
A= xz/1+xz+z + 1/z+1+xz + x/xz+z+1 = xz+1+x/xz+1+x =1
Vay A=1
Thay xyz = 2011 vào N được :
\(N=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}=\frac{xy.xz}{xy\left(z+xz+1\right)}+\frac{y}{y\left(z+xz+1\right)}+\frac{z}{z+xz+1}\)
\(=\frac{xz}{z+xz+1}+\frac{1}{z+xz+1}+\frac{z}{z+xz+1}=\frac{z+xz+1}{z+xz+1}=1\)
Ta có: \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Leftrightarrow\left(x+y\right)^2=\left(-z\right)^2\)
\(\Leftrightarrow x^2+2xy+y^2=z^2\)
\(\Leftrightarrow x^2+y^2-z^2=-2xy\)
Chứng minh tương tự ta có:
\(x^2+z^2-y^2=-2xz\)
\(y^2+z^2-x^2=-2yz\)
\(\frac{xy}{x^2+y^2-z^2}+\frac{xz}{x^2+z^2-y^2}+\frac{yz}{y^2+z^2-x^2}\)
\(=\frac{xy}{-2xy}+\frac{xz}{-2xz}+\frac{yz}{-2yz}\)
\(=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\)
\(=-\frac{3}{2}\)
Vậy giá trị biểu thức là \(-\frac{3}{2}\)
với xyz=2009, thay vào, ta có
\(A=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
=\(\frac{xz}{1+zx+y}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}=1\)
=> ... k phụ thuộc vào x,y,z(ĐPCM)
^_^
Bạn tham khảo tại đây:
Câu hỏi của trieu dang - Toán lớp 8 - Học toán với OnlineMath
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{\left(yz+xz+xy\right)}{xyz}=0\)
\(\Rightarrow yz+zx+xy=0\)
Ta có : \(x^2+2yz=x^2+yz+yz\)
\(=x^2+yz-zx-xy\)
\(=x\left(x-z\right)-y\left(x-z\right)\)
\(=\left(x-y\right)\left(x-z\right)\)
Tương tự : \(y^2+2xz=y^2+xz+xz\)
\(=y^2+xz-xy-yz\)
\(=y\left(y-x\right)+z\left(x-y\right)\)
\(=\left(x-y\right)\left(z-y\right)\)
\(z^2+2xy=\left(x-z\right)\left(y-z\right)\)
\(\Rightarrow M=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\) \(M=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)
\(M=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\frac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
\(M=\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}\)
Vì xyz=1 nên \(x\ne0;y\ne0;z\ne0\)
Ta có \(\frac{1}{1+x+xy}=\frac{z}{\left(1+y+yz\right)xz}=\frac{xz}{z+xz+1}\)
Tương tự \(\frac{1}{1+y+yz}=\frac{xz}{\left(1+y+yz\right)xz}=\frac{xz}{xz+z+1}\)
Khi đó \(M=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{1+z+xz}=\frac{z+xz+1}{z+zx+1}=1\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{xy+yz+zx}{xyz}=0\Leftrightarrow xy+yz+zx=0\)
\(\Leftrightarrow xy=-yz-zx;yz=-xy-zx;zx=-xy-yz\)
Ta có: x2+2yz=x2+yz+yz=x2+yz-xy-zx=x(x-y)-z(x-y)=(x-y)(x-z)
Tương tự: \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2+2xy=\left(z-x\right)\left(z-y\right)\)
A= \(\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}\)=\(\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)
\(=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)\(=\frac{xy\left(x-y\right)-xz\left(x-y+y-z\right)+yz\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\frac{xy\left(x-y\right)-xz\left(x-y\right)-xz\left(y-z\right)+yz\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)\(=\frac{\left(xy-xz\right)\left(x-y\right)-\left(xz-yz\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\frac{x\left(y-z\right)\left(x-y\right)-z\left(x-y\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=1\)
\(\hept{\begin{cases}xyz=12\\x^3+y^3+z^3=36\end{cases}}\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)-3xyz+z^3=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow x=y=z\left(x+y+z>0\right)\)
Thay x=y=z vào r tính thôi bạn