Ta có: \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0\)

=>\(\frac{yz + zx + xy}{xyz}=0\)

=>xy+yz+xz=0

=>yz=-xy-xz; xy=-xz-yz; xz=-xy-yz

\(x^2 + 2yz = x^2 + yz + yz = x^2 - x(y + z) + yz\)

\(=x^2-xy-xz+yz\)

\(=x(x-y)-z(x-y)=(x-y)(x-z)\)

Chứng minh tương tự, ta sẽ có:

\(y^2 + 2zx = (y - z)(y - x)\)

\(z^2 + 2xy = (z - x)(z - y)\)

Đặt \(A = \frac{1}{x^2 + 2yz} + \frac{1}{y^2 + 2zx} + \frac{1}{z^2 + 2xy}\)

\(=\frac{1}{(x - y)(x - z)}+\frac{1}{(y - z)(y - x)}+\frac{1}{(z - x)(z - y)}\)

\(=\frac{-1}{(x - y)(z - x)}+\frac{-1}{(y - z)(x - y)}+\frac{-1}{(z - x)(y - z)}\)

\(=\frac{-(y - z) - (z - x) - (x - y)}{(x - y)(y - z)(z - x)}\)

=0

=>\(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}=0\)

\(\left(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}\right)\left(x^{2016}+y^{2017}+z^{2018}\right)\)

\(=0\left(x^{2016}+y^{2017}+z^{2018}\right)\)

=0

=xy+yz+xz

cũng bằng 3

Ta coˊ :xy+x+1x​+yz+y+1y​+xz+z+1z​

=���+�+1+�����+��+�+����2��+���+��=xy+x+1x​+xyz+xy+xxy​+x2yz+xyz+xyxyz​

=���+�+1+����+�+1+1��+�+1(Vıˋ ���=1)=xy+x+1x​+xy+x+1xy​+xy+x+11​(V

Ta có: \(x+y+z=0\)

=> \(\left(x+y+z\right)^2=0\)

<=> \(x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)

<=> \(x^2+y^2+z^2=0\) ( Dô \(xy+yz+xz=0\) )

=> \(x=y=z=0\) (1)

Thay (1) vào Q ta được:

Q = \(\left(-1\right)^{2017}+0^{2018}+1^{2019}=0\)

hơi dài mà lười nên mình nói cách làm nha :P

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow xy+yz+xz=0\)

bạn cm \(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}=0\)

tách: \(x^2+2yz=x^2+yz-xy-xz=\left(x-z\right).\left(x-y\right)\), mấy cái khác tương tự 

quy đồng rồi tính ra = 0 là được 

Vì \(17.\left(xy+yz+zx\right)=105\Rightarrow\left(xy+yz+zx\right)=\frac{105}{17}\)

Ta có :

\(\left(x+z+y\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)=19+2\left(\frac{105}{17}\right)=31\frac{6}{17}\)

Do đó : \(x+y+z=\sqrt{31\frac{6}{17}}\)

hoặc \(x+y+z=-\sqrt{31\frac{6}{17}}\) 

Chúc bạn học tốt nha !!!

Lời giải:

\(yz-xz-xy=0\Rightarrow yz-xz=xy\)

\(B=\frac{yz}{x^2}-\frac{zx}{y^2}-\frac{xy}{z^2}\)\(=\frac{(yz)^3-(xz)^3-(xy)^3}{x^2y^2z^2}\)

Xét: \((yz)^3-(xz)^3-(xy)^3=(yz-xz)^3+3yz.xz(yz-xz)-(xy)^3\)

\(=(xy)^3+3yz.xz.xy-(xy)^3=3x^2y^2z^2\)

\(\Rightarrow B=\frac{(yz)^3-(xz)^3-(xy)^3}{x^2y^2z^2}=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)

Từ xy+yz+zx=0 => 2(xy+yz+zx)=0

Từ x+y+z=0 => (x+y+z)²=0

=>x²+y²+z²+2(xy+yz+zx)=0

=>x²+y²+z²=0

Mà \(x^2\ge0,y^2\ge0,z^2\ge0\Rightarrow x^2+y^2+z^2\ge0\)

=>x=y=z=0

Thay x=y=z=0 vào A

=>A=-1+1-1=-1

nhàm dòng cuối 

=>A=-1+0+1=0