Ta có: \(\left(\sqrt{x+y}\right)^2=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

\(\Leftrightarrow\)\(x+y=x+y-2z+2\sqrt{\left(x-z\right)\left(y-z\right)}\)

\(\Leftrightarrow2z=2\sqrt{\left(x-z\right)\left(y-z\right)}\)

Theo giả thiết, ta có: 

theo giả thiết, ta có: \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Rightarrow\frac{1}{z}-\frac{1}{x}=\frac{1}{y}\)\(\Rightarrow\frac{x-z}{zx}=\frac{1}{y}\Rightarrow x-z=\frac{zx}{y}\)

Tương tự, ta có: \(y-z=\frac{zy}{x}\)

Do đó: \(2\sqrt{\left(x-z\right)\left(y-z\right)}=2\sqrt{\frac{zx}{y}.\frac{zy}{x}}=2z\) (1)

ta có: \(\left(\sqrt{x+y}\right)^2=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

\(\Leftrightarrow2z=2\sqrt{\left(x-z\right)\left(y-z\right)}\)(2)

Thay (2) vào (1) ta thấy (2) luôn đúng

Suy ra ĐPCM

Vì \(x>0,y>0\Rightarrow\frac{1}{x}>0;\frac{1}{y}>0\)

mà \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Rightarrow\frac{1}{z}=\frac{1}{x}+\frac{1}{y}\Rightarrow\frac{1}{z}>0\Rightarrow z>0\)

Ta có: \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Leftrightarrow yz+zx-xy=0\)

\(\Leftrightarrow-z^2=-z^2+yz+zx-xy=-\left(x-z\right)\left(y-z\right)\)

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

\(\Rightarrow z=\sqrt{\left(x-z\right)\left(y-z\right)}\left(z>0\right)\)

Lại có: \(x+y=x-z+y-z+2z\)

\(=\left(x-z\right)+\left(y-z\right)+2\sqrt{\left(x-z\right)\left(y-z\right)}=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

Suy ra \(\sqrt{x+y}=\sqrt{x-z}+\sqrt{y-z}\) (ĐPCM)

`Answer:`

\(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{zx}{cx+ax}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(1\right)\)

Theo đề ra, có: \(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{zx}{cx+az}\)

\(\Rightarrow\frac{xyz}{ayz+bxz}=\frac{xyz}{bxz+cxy}=\frac{xyz}{cxy+ayz}\)

\(\Rightarrow ayz+bxz=bxz+cxy=cxy+ayz\)

\(\Rightarrow\hept{\begin{cases}ayz+bxz=bxz+cxy\\ayz+bxz=cxy+ayz\\bxz+cxy=cxy+ayz\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}ayz=cxy\\bxz=cxy\\bxz=ayz\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}az=cx\\bz=cy\\bx=ay\end{cases}}\left(2\right)\)

Thế (2) và (1): \(\frac{xy}{2ay}=\frac{yz}{2bz}=\frac{xz}{2cx}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

\(\Rightarrow\frac{x}{2a}=\frac{y}{2b}=\frac{z}{2c}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(3\right)\)

\(\Rightarrow\frac{x^2}{4a^2}=\frac{y^2}{4b^2}=\frac{z^2}{4c^2}=\frac{\left(x^2+y^2+z^2\right)^2}{\left(a^2+b^2+c^2\right)^2}=\frac{x^2+y^2+z^2}{4a^2+4b^2+4c^2}\)

\(\Rightarrow\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{1}{4}\)

Thế (3) vào (2): \(\frac{x}{2a}=\frac{y}{2b}=\frac{z}{2c}=\frac{1}{4}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{a}{2}\\y=\frac{b}{2}\\z=\frac{c}{2}\end{cases}}\)

\(\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{x+z}{y}=\dfrac{x^2y+xy^2+y^2z+yz^2+x^2z+xz^2}{xyz}=\dfrac{-3xyz}{xyz}=-3\)

đề cho xy+yz+xz=0 nhân cả 2 vế với -z

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

=>-xyz=\(z^2x+z^2y\)

cmtt bạn nhân với -y và -z

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

sửa đề: z+4>0

Đặt a = x + 1 > 0 ; b = y + 1 > 0 ; c = z + 4 > 0

a + b + c = 6

\(A=\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-4}{c}=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)

Theo Bất Đẳng Thức ta có: \(\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{4}{c}\ge\frac{4}{a+b}+\frac{4}{c}\ge\frac{16}{a+b+c}=\frac{8}{3}\)

\(\Rightarrow A\le\frac{1}{3}\)Đẳng thức xảy ra khi và chỉ khi \(\hept{\begin{cases}a=b\\a+b=c\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b=\frac{3}{2}\\c=3\end{cases}\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{2}\\z=-1\end{cases}}}\)

Vậy MaxA = 1/3 khi \(\hept{\begin{cases}x=y=\frac{1}{2}\\z=-1\end{cases}}\)

PaiN: Nhưng x,y,z là các số thực dương thì sao z âm đc?

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

\(\frac{x^3}{z^2+x^2}=x-\frac{z^2x}{z^2+x^2}\ge x-\frac{z^2x}{2zx}=x-\frac{z}{2}\)

\(\frac{y^3}{y^2+z^2}=y-\frac{yz^2}{y^2+z^2}\ge y-\frac{yz^2}{2yz}=y-\frac{z}{2}\)

\(\frac{x^2+y^2+4}{x+y}=\frac{z\left(x+y\right)+4}{x+y}=z-x-y+\frac{4}{x+y}+x+y\ge z-x-y+4\)

Cộng lại ra minP=4, dấu "=" xảy ra khi \(x=y=z=1\)