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​(Vıˋ xyz=1)

=�+��+1��+�+1=xy+x+1x+xy+1​

$=1$

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

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

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

nhầm xíu nhá mk lm lại :

\(A=\frac{xz}{z\left(xy+x+1\right)}+\frac{xyz}{xz\left(yz+y+1\right)}+\frac{z}{xz+z+1}\)\(=\frac{xz}{xyz+xz+z}+\frac{1}{xyz^2+xyz+xz}+\frac{z}{xz+z+1}\)

\(=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\)

\(=\frac{xz+z+1}{xz+z+1}=1\)

\(A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}=\frac{xz}{z\left(xy+x+1\right)}+\frac{xyz}{xz\left(yz+y+1\right)}+\frac{z}{xz+z+1}\)

\(=\frac{xy}{xyz+xz+z}+\frac{1}{xyz^2+xyz+xz}+\frac{z}{xz+z+1}=\frac{xy}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

\(=\frac{xy+1+z}{xz+z+1}=1\)

vậy A=1

\(\text{Ta có :}\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\)

\(=\frac{x}{xy+x+1}+\frac{xy}{xyz+xy+x}+\frac{xyz}{x^2yz+xyz+xy}\)

\(=\frac{x}{xy+x+1}+\frac{xy}{xy+x+1}+\frac{1}{xy+x+1}\left(\text{Vì }xyz=1\right)\)

\(=\frac{x+xy+1}{xy+x+1}\)

\(=1\)

Áp dụng bất đẳng thức Bunyakovsky:

\(P^2=\left(\sqrt{2x+yz}+\sqrt{2y+xz}+\sqrt{2z+xy}\right)^2\)

\(\le\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)\)

\(=3\left(4+xy+yz+xz\right)=12+3\left(xy+yz+xz\right)\)

Mặt khác,theo AM-GM:

\(3\left(xy+yz+xz\right)\le\left(x+y+z\right)^2=4\)

\(\Rightarrow12+3\left(xy+yz+xz\right)\le12+4=16\)

\(\Rightarrow P^2\le16\Leftrightarrow P\le4\)

Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)