Có : \(x-2y-\sqrt{xy}+\sqrt{x}-2\sqrt{y}=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{x}-2\sqrt{y}=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+1\right)=0\)

\(\Leftrightarrow\sqrt{x}=2\sqrt{y}\) (Do \(\sqrt{x}+\sqrt{y}+1>0,\forall x;y>0\))

\(\Leftrightarrow x=4y\)

Khi đó \(P=\dfrac{7y}{\left(2\sqrt{y}+3\sqrt{y}\right).\left(\sqrt{x}+2\sqrt{y}\right)}\)

\(=\dfrac{7y}{5\sqrt{y}.4\sqrt{y}}=\dfrac{7}{20}\)

\(VT=\sum\sqrt{\frac{1}{2}\left(x^2+2xy+y^2\right)+\frac{3}{2}\left(x^2+y^2\right)}\)

\(VT\ge\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{3}{4}\left(x+y\right)^2}=\sum\sqrt{\frac{5}{4}\left(x+y\right)^2}\)

\(VT\ge\frac{\sqrt{5}}{2}\left(x+y\right)+\frac{\sqrt{5}}{2}\left(y+z\right)+\frac{\sqrt{5}}{2}\left(z+x\right)\)

\(VT\ge\sqrt{5}\left(x+y+z\right)=\sqrt{5}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)

1.

\(\left\{{}\begin{matrix}x-2y-\sqrt{xy}=0\\\sqrt{x-1}-\sqrt{2y-1}=1\end{matrix}\right.\)

\(pt\left(1\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-2\sqrt{y}\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=-\sqrt{y}\\\sqrt{x}=\sqrt{2y}\end{matrix}\right.\)

cái đầu tiên loại vì x=y=0 không phải là nghiệm của hệ

suy ra x=2y thày vào pt(2) ta thấy 0 = 1 vô lý

vậy pt vô nghiệm

Phân tích cái trên thành hằng đẳng thức bậc 2 là đc, tìm ra x;y;z rồi thay vào M

làm rõ ra cho tớ được không? Không hiểu sao tớ phân tích không ra :((

a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)

b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)

c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)

(4x + 2y + 2z - \(\sqrt{4xy}-\sqrt{4xz}+2\sqrt{yz}\) )+(y - \(6\sqrt{y}\) + 9)+(z- \(10\sqrt{z}\) + 25) = 0

<=> (\(2\sqrt{x}-\sqrt{y}-\sqrt{z}\))² + (\(\sqrt{y}-3\))² + (\(\sqrt{z}-5\))² = 0 (1)

Vì VP \(\ge0\) => để (1) có n₀ thì

\(\left\{{}\begin{matrix}2\sqrt{x}-\sqrt{y}-\sqrt{z}=0\left(x\right)\\\sqrt{y}-3=0\left(xx\right)\\\sqrt{z}-5=0\left(xxx\right)\end{matrix}\right.\)

Từ(xx) => \(\sqrt{y}=3\) <=> y = 9

Từ (xxx) => \(\sqrt{z}=5\) <=> z = 25

Từ (x) => \(2\sqrt{x}=8\) <=> \(\sqrt{x}=4\) <=> x = 16

=> M = (16 - 15)² + (9 - 8)² + (25 - 24)² = 1 + 1 + 1 = 3

Dạo này ko tag được đâu :(

\(VT=\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{1}{2}\left(x^2+y^2\right)+y^2}\ge\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{1}{4}\left(x+y\right)^2+y^2}\)

\(VT\ge\sum\sqrt{\frac{3}{4}\left(x+y\right)^2+y^2}\ge\sqrt{\frac{3}{4}\left(2x+2y+2z\right)^2+\left(x+y+z\right)^2}\)

(Mincopxki)

\(\Rightarrow VT\ge\sqrt{4\left(x+y+z\right)^2}=2\left(x+y+z\right)\)

@Nguyễn Việt Lâm

\(\hept{\begin{cases}3x^2-2y^2-xy+12x-17y-15=0\left(1\right)\\\sqrt{2-x}+\sqrt{6-x-x^2}=y+\sqrt{2y+5}-\sqrt{y+4}\left(2\right)\end{cases}}\)

PT (1) \(\Leftrightarrow3x^2-x\left(y-12\right)-2y^2-17y-15=0\)

\(\Leftrightarrow\Delta=\left(y-12\right)^2+4\cdot3\cdot\left(2y^2+17y+15\right)\)

\(\Leftrightarrow\Delta=y^2-24y+144+24y^2+204y+180\)

\(\Leftrightarrow\Delta=25y^2+180y+324\)

\(\Delta=\left(5y+18\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{y-12+5y+18}{3}=2y+2\\x=\frac{y-12-5y-18}{3}=\frac{-4y}{3}-10\end{cases}}\)

\(x=2y+2\)

\(\Leftrightarrow\sqrt{2-x}+\sqrt{6-x-x^2}=y+\sqrt{2y+5}-\sqrt{y+4}\)

\(\Leftrightarrow\sqrt{-2y}+\sqrt{6-2y-2-4y^2-8y-4}=y+\sqrt{2y+5}-\sqrt{y+4}\)

\(\Leftrightarrow\sqrt{-2y}+\sqrt{-4y^2-10y+0}=y+\sqrt{2y+5}-\sqrt{y+6}\)

\(\Leftrightarrow y=0\Rightarrow x=2\)

Vậy (x;y)=(2;0)