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x^2+y^2+z^2/y^2-2yx+z^2+z^2-2xy+x^2+x^2-2xy+y^2=x^2+y^2+z^2/2y^2+2x^2+2z^2-6xy=x^2+y^2+z^2/2(x^2+y^2+z^2)-6xy=1/2-6xy
xét mẫu ta có
=y^2 - 2yz + z^2 + z^2 -2xz + x^2 + x^2 -2xy +y^2
thêm bớt x^2,y^2,z^2 vào mẫu ta có
=3y^2 + 3x^2 + 3z^2 - (x^2 + y^2 + z^2 + 2xy + 2yz + 2xz)
đúng không
mà (x+y+z)=0 => (x+y+z)^2=0
mà (x^2 + y^2 + z^2 + 2xy + 2yz + 2xz) phân tích ra thành (x+y+z)^2
=> (x^2 + y^2 + z^2 + 2xy + 2yz + 2xz)=0
=> (x^2 + y^2 + z^2 )/ 3(x^2 + y^2 + z^2)
rút gọn thành 1/3
nhớ k nha chuẩn 100%
\(=\frac{x^2+y^2+z^2}{2x^2+2y^2+2z^2-2xy-2yz-2zx}=\frac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x^2+y^2+z^2+2xy+2yz+2xz\right)}\)
\(=\frac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2}=\frac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\frac{1}{3}\)(vì x+y+z=0)
tách mẫu số ra được: 2(x2+y2+z2)-2(xy+yz+xz) (1)
mà x+y+z=0
=> (x+y+z)2=0
=> x2+y2+z2= -2(xy+yz +xz) (2)
Thay (2) vào (1) ta được mẫu số: 3(x2+y2+z2)
Phân thức khi rút gọn được là: 1/3
\(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\frac{\left(x+y+z\right)^2-2\left(xy+yz+xz\right)}{2x^2+2y^2+2z^2-2xy+2yz+2xz}\)
\(=\frac{-2\left(xy+yz+xz\right)}{2\left(x+y+z\right)^2-6\left(xy+yz+xz\right)}\)
\(=-\frac{1}{3}\)
Có:\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right) ^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(\Rightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\)
Có:
\(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\)
\(=y^2-2yz+z^2+z^2-2xz+z^2+x^2-2xy+y^{^2}\)
\(=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)
\(=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2\)
\(=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}\)
\(=\dfrac{1}{3}\)
Lời giải:
Đặt \(B=\frac{x^2+y^2+z^2}{(y-z)^2+(z-x)^2+(x-y)^2}\)
\(\Leftrightarrow B=\frac{x^2+y^2+z^2}{2(x^2+y^2+z^2-xy-yz-xz)}\)
Lại có
\(x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)\)
\(\Rightarrow B=\frac{-2(xy+yz+xz)}{2[-2(xy+yz+xz)-(xy+yz+xz)]}=\frac{-2(xy+yz+xz)}{-6(xy+yz+xz)}=\frac{1}{3}\)
Ta có: \(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(x^2+y^2+z^2+2xy+2yz+2xz=0\)
\(\Rightarrow x^2+y^2+z^2=-2.\left(xy+yz+zx\right)\)
\(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\frac{-2.\left(xy+yz+zx\right)}{y^2+z^2+z^2+x^2+x^2+y^2-2.\left(xy+yz+zx\right)}\)
\(=\frac{-2.\left(xy+yz+zx\right)}{2.\left(x^2+y^2+z^2\right)-2.\left(xy+yz+zx\right)}\)
\(=\frac{-2.\left(xy+yz+zx\right)}{2.\left[-2.\left(xy+yz+zx\right)\right]-2.\left(xy+yz+zx\right)}\)
\(=\frac{-2.\left(xy+yz+zx\right)}{-6.\left(xy+yz+zx\right)}\)
\(=\frac{1}{3}\left(xy+yz+zx\ne0\right)\)
Tham khảo nhé~
\(A=\dfrac{bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2bcyz-2cazx-2abxy}{ax^2+by^2+cz^2}=\dfrac{\left(bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\right)-\left(ax+by+cz\right)^2}{ax^2+by^2+cz^2}=\dfrac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)
\(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2=-2\left(xy+yz+zx\right)\)
\(P=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)}=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2}=\dfrac{1}{3}\)