\(\left(x\ne-y;x>\dfrac{y}{2}\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7-\left(x+y\right)}{x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7}{x+y}=2\end{matrix}\right.\)

\(đặt:\dfrac{1}{\sqrt{2x-y}}=a>0;\dfrac{1}{x+y}=b\)

\(\Rightarrow\left\{{}\begin{matrix}4a-21b=\dfrac{1}{2}\\3a+7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\left(tm\right)\\b=\dfrac{1}{14}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x-y}}=\dfrac{1}{2}\\\dfrac{1}{x+y}=\dfrac{1}{14}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)(thỏa)

1: \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{y}=3\\\left|x-1\right|=2-\dfrac{2}{y}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=2-\dfrac{2}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{2;0\right\}\end{matrix}\right.\)

2: \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\2\left|x-1\right|+\dfrac{4}{y-1}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{9}{y-1}=-9\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=3-\dfrac{2}{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{3;-1\right\}\end{matrix}\right.\)

3: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)

a.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)

b.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)

(1) + rút y từ pt (2) thay vào pt (1), ta được pt bậc hai 1 ẩn x, dễ rồi, tìm x rồi suy ra y

(2) + (3)

+ pt nào có nhân tử chung thì đặt nhân tử chung (thật ra chỉ có pt (2) của câu 2 là có nhân từ chung)

+ trong hệ, thấy biểu thức nào giống nhau thì đặt cho nó 1 ẩn phụ

VD hệ phương trình 3: đặt a= x+y ; b= căn (x+1)

+ khi đó ta nhận được một hệ phương trình bậc nhất hai ẩn, giải hpt đó rồi suy ra x và y

1. Với mọi số thực x;y;z ta có:

\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)

\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)

\(\Rightarrow P\ge3\)

\(P_{min}=3\) khi \(x=y=z=1\)

1.1

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)

\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)

\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)

\(\Leftrightarrow a=b\Leftrightarrow x=y\)

Thay vào pt đầu:

\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))

\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)

\(\Rightarrow a=1\Rightarrow x=y=1\)

2.

\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)

\(\Rightarrow4x^2-10xy+4y^2=0\)

\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)

Thế vào pt đầu

...

Ta có: \(\left\{{}\begin{matrix}\dfrac{3}{\sqrt{x+y}}-\dfrac{2}{\sqrt{x-y}}=4\\\dfrac{2}{\sqrt{x+y}}-\dfrac{1}{\sqrt{x-y}}=5\end{matrix}\right.\)

Đặt: \(t=\sqrt{x+y}\) và \(k=\sqrt{x-y}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{t}-\dfrac{2}{k}=4\\\dfrac{2}{t}+\dfrac{1}{k}=5\end{matrix}\right.\)

Ta lại đặt: \(a=\dfrac{1}{t}\) và \(u=\dfrac{1}{k}\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a-2u=4\\2a+u=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a-2u=4\\4a+2u=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a-2u=4\\7a=14\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6-2u=4\\a=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u=1\\a=2\end{matrix}\right.\)

Mà: 

\(u=1\Rightarrow\dfrac{1}{k}=1\Rightarrow k=1\)

\(a=2\Rightarrow\dfrac{1}{t}=2\Rightarrow t=\dfrac{1}{2}\)

Ta lại có:

\(k=1\Rightarrow\sqrt{x+y}=1\)

\(t=\dfrac{1}{2}\Rightarrow\sqrt{x-y}=\dfrac{1}{2}\)

Ta có hệ:

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\x+y=\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\2x=\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{8}-y=1\\x=\dfrac{5}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{8}\\x=\dfrac{5}{8}\end{matrix}\right.\)

Vậy \(x-\dfrac{5}{8};y=-\dfrac{3}{8}\)

Đặt 1/căn x+y=a; 1/căn x-y=b

Theo đề, ta có hệ:

3a-2b=4 và 2a+b=5

=>a=2 và b=1

=>x+y=1/4 và x-y=1

=>x=5/8 và y=-3/8