\(A=\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

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

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

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

\(A=\dfrac{yz}{y+1+yz}+\dfrac{1}{y+yz+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz+1}{y+1+yz}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz+xyz}{y+xyz+yz}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{y\left(z+xz\right)}{y\left(1+xz+z\right)}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{z+xz+1}{xz+z+1}\)

\(A=1\)

\(A=\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)⇔\(A=\dfrac{z}{1+xz+z}+\dfrac{xz}{z+1+xz}+\dfrac{1}{xz+z+1}\)(vì xyz=1)

⇔\(A=\dfrac{z+xz+1}{xz+z+1}\)⇔\(A=1\)

Xong rồi nè bn ơi 

\(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(=\dfrac{1}{\dfrac{1}{z}+\dfrac{1}{yz}+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{\dfrac{1}{y}+z+1}\)

\(=\dfrac{1}{\dfrac{y+1+yz}{yz}}+\dfrac{1}{yz+y+1}+\dfrac{1}{\dfrac{1+zy+y}{y}}\)

\(=\dfrac{yz}{y+1+yz}+\dfrac{1}{yz+y+1}+\dfrac{y}{1+zy+y}=\dfrac{y+yz+1}{y+yz+1}=1\)

b, đk: \(x\ge1,y\ge2,z\ge3\)

\(=>B=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)

đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{y-2}=b\\\sqrt{z-3}=c\end{matrix}\right.\)\(=>\left\{{}\begin{matrix}x=a^2+1\\y=b^2+1\\z=c^2+1\end{matrix}\right.\)\(=>a\ge0,b\ge0,c\ge0\)

B trở thành \(\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\)

\(=\dfrac{a^{ }}{a^2+1}+\dfrac{a^2+1}{4}+\dfrac{b}{b^2+1}+\dfrac{b^2+1}{4}+\dfrac{c}{c^2+1}+\dfrac{c^2+1}{4}\)

\(-\left(\dfrac{a^2+b^2+c^2+3}{4}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}-\dfrac{a^2+b^2+c^2}{4}\)\(=0\)

dấu"=" xảy ra<=>\(a=0,b=0,c=0< =>x=1,y=2,z=3\)

Chắc bạn ghi nhầm đề, tìm GTLN mới đúng, chứ GTNN của các biểu thức này đều hiển nhiên bằng 0

\(A=\dfrac{3.\sqrt{x-9}}{15x}\le\dfrac{3^2+x-9}{30x}=\dfrac{1}{30}\)

\(A_{max}=\dfrac{1}{30}\) khi \(x=18\)

\(B=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}=\dfrac{1.\sqrt{x-1}}{x}+\dfrac{\sqrt{2}.\sqrt{y-2}}{\sqrt{2}y}+\dfrac{\sqrt{3}.\sqrt{z-3}}{\sqrt{3}z}\)

\(B\le\dfrac{1+x-1}{2x}+\dfrac{2+y-2}{2\sqrt{2}y}+\dfrac{3+z-3}{2\sqrt{3}z}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;4;6\right)\)

đề bài là tìm gt lớn nhất nhé mọi người,tớ ghi nhầm

2 cái kìa còn lại làm tương tự rồi sau đó cộng lại với nhau sẽ ra 1 số tự nhiên nhé, dễ nên lười đánh nốt lắm :v

cam ơn ah. kết quả bằng 3 ah.

\(A=\dfrac{xyz.x}{xy+xyz.x+xyz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{xyz+yz+y}\)

\(=\dfrac{xz}{1+xz+z}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)

\(=\dfrac{xyz}{y+xyz+yz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)

\(=\dfrac{2019}{y+2019+yz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)

\(=\dfrac{yz+y+2019}{yz+y+2019}=1\)

Đề thiếu kìa :vv

⇔xy+yz+zx=0

=yz/(x−y)(x−z)

Tương tự: xy/z^2+2xy=xy/(x−z)(y−z)

\(\dfrac{1}{\left(x-y\right)\left(z^2+yz-x^2-xz\right)}=\dfrac{1}{\left(x-y\right)\left[\left(z-x\right)\left(z+x\right)+y\left(z-x\right)\right]}=\dfrac{1}{\left(z-x\right)\left(x-y\right)\left(x+y+z\right)}\)

Tương tự: \(\dfrac{1}{\left(y-z\right)\left(x^2+xz-y^2-yz\right)}=\dfrac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}\)

\(\dfrac{1}{\left(z-x\right)\left(y^2+xy-z^2-xz\right)}=\dfrac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}\)

\(\Rightarrow M=\dfrac{y-z-z+x-x+y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}\\ M=\dfrac{2}{\left(x-y\right)\left(z-x\right)\left(x+y+z\right)}\)

\(a,\) Bổ sung điều kiện: \(x\ne y\ne z\)

\(b,\) Đề bài ko thỏa mãn điều kiện nên không tính đc M

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)

\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)

Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)

\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)