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

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

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

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

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

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mn giúp mình với

nếu khó nhìn để mik đánh lại

Ta có:\(A=\dfrac{xy}{x+y}+\dfrac{yz}{y+z}+\dfrac{zx}{z+x}\)

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

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

Ta có:\(\dfrac{x^2}{x+y}+\dfrac{x+y}{9}\ge2\sqrt{\dfrac{x^2}{x+y}.\dfrac{x+y}{9}}=\dfrac{2x}{3}\)

Tương tự,ta có:\(\dfrac{y^2}{y+z}+\dfrac{y+z}{9}\ge\dfrac{2y}{3};\dfrac{z^2}{z+x}+\dfrac{z+x}{9}\ge\dfrac{2z}{3}\)

Cộng vế với vế ta có:

\(\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}+\dfrac{2\left(x+y+z\right)}{4}\ge\dfrac{2\left(x+y+z\right)}{3}\)

\(\Leftrightarrow\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}\ge\dfrac{2\left(x+y+z\right)}{3}-\dfrac{2\left(x+y+z\right)}{4}=\dfrac{2.9}{3}-\dfrac{9}{2}=\dfrac{3}{2}\)

\(\Rightarrow A\le9-\dfrac{3}{2}=\dfrac{15}{2}\)

Dấu "=" xảy ra ⇔ x=y=z=3

Vậy,Max A=\(\dfrac{15}{2}\) ⇔ x=y=z=3

Xét \(x+y+z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+z=-x\\z+x=-y\\x+y=-z\end{matrix}\right.\)

\(\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1\)

Xét \(x+y+z\ne0\) thì ta có:

\(\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\\5y=z+x+3y\\5z=x+y+3z\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\\2y=z+x\\2z=x+y\end{matrix}\right.\)

\(\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64\)

Vậy \(\left[{}\begin{matrix}A=1\\A=64\end{matrix}\right.\)

Nếu bị lỗi thì bạn có thể xem đây nhé:

Áp dụng bđt AM-GM ta có:

\(\frac{x^2}{x+y}+\frac{x+y}{4}\ge2\sqrt{\frac{x^2}{x+y}.\frac{x+y}{4}}=x\)

\(\frac{y^2}{x+z}+\frac{x+z}{4}\ge2\sqrt{\frac{y^2}{x+z}.\frac{x+z}{4}}\ge y\)

\(\frac{z^2}{x+y}+\frac{x+y}{4}\ge2\sqrt{\frac{z^2}{x+y}.\frac{x+y}{4}}\ge z\)

Cộng từng vế các bđt trên ta được:

\(P+\frac{x+y+z}{2}\ge x+y+z\)

\(\Rightarrow P\ge\frac{x+y+z}{2}=1\)

Dấu"="xảy ra \(\Leftrightarrow x=y=z=1\)

Vậy Min P=1 \(\Leftrightarrow x=y=z=1\)

anh Châu ơi, 1+1+1 đâu có = 2 anh.

à anh xl nhầm x=y=z=\(\frac{2}{3}\)

Lời giải:
$2(x+y)=3(y+z)=4(x+z)$

$\Rightarrow \frac{x+y}{6}=\frac{y+z}{4}=\frac{x+z}{3}$ (chia cả 3 vế cho $12$)

Đặt giá trị trên là $t$

$\Rightarrow x+y=6t; y+z=4t; z+x=3t$

$\Rightarrow x+y+z=(6t+4t+3t):2=6,5t$

$x=6,5t-4t=2,5t; y=6,5t-3t=3,5t; z=6,5t-6t=0,5t$. Khi đó:
$P=\frac{2,5t}{3,5t}+\frac{3,5t}{0,5t}+\frac{0,5t}{2,5t}$

$=\frac{2,5}{3,5}+\frac{3,5}{0,5}+\frac{0,5}{2,5}=\frac{277}{35}$