1.

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

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

Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=-\dfrac{5}{4}\\a^2+b=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-a^2-\dfrac{5}{4}-a\left(a^2+\dfrac{5}{4}\right)=-\dfrac{5}{4}\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a^3-\dfrac{1}{4}a=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-a\left(a^2-a+\dfrac{1}{4}\right)=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a-\dfrac{1}{2}\right)^2=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=0\\xy=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt[3]{10}}{2}\\y=-\dfrac{5}{2\sqrt[3]{10}}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=\dfrac{1}{2}\\xy=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)

Kết luận: Phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{\sqrt[3]{10}}{2};-\dfrac{5}{2\sqrt[3]{10}}\right);\left(1;-\dfrac{3}{2}\right)\right\}\)

2.

\(\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\left(\dfrac{2}{y}\right)^3-4\left(\dfrac{2}{y}\right)\\1+\left(\dfrac{2}{y}\right)^2=5\left(x+1\right)^2+5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=u\\\dfrac{2}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-16u=v^3-4v\\v^2=5u^2+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u^3-v^3=16u-4v\\4=v^2-5u^2\end{matrix}\right.\)

\(\Rightarrow4\left(u^3-v^3\right)=\left(16u-4v\right)\left(v^2-5u^2\right)\)

\(\Leftrightarrow21u^3-5u^2v-4uv^2=0\)

\(\Leftrightarrow u\left(7u-4v\right)\left(3u+v\right)=0\Rightarrow\left[{}\begin{matrix}u=0\Rightarrow v^2=4\\u=\dfrac{4v}{7}\Rightarrow4=v^2-5\left(\dfrac{4v}{7}\right)^2\\v=-3u\Rightarrow4=\left(-3u\right)^2-5u^2\end{matrix}\right.\) 

\(\Rightarrow...\)

HPT đã cho

\(\left\{{}\begin{matrix}\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}=\dfrac{2}{3}\\\dfrac{x^2+1}{x}+\dfrac{y^2+1}{y}=6\end{matrix}\right.\)

Đặt \(\dfrac{x}{x^2+1}=u;\dfrac{y}{y^2+1}=v\)

HPT tương đương

\(\left\{{}\begin{matrix}u+v=\dfrac{2}{3}\\\dfrac{1}{u}+\dfrac{1}{v}=6\end{matrix}\right.\)

Tới đây thì dễ rồi u=1/3;v=1/3

Xong tìm được x,y

a.

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)

\(\Rightarrow x^2+\left(\dfrac{2}{x}\right)^2=5\)

\(\Leftrightarrow x^4-5x^2=4=0\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: ...

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}\right)^2-\left(y+\dfrac{1}{y}\right)^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)\left(x+\dfrac{1}{x}-y-\dfrac{1}{y}\right)=21\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+1=0\\y^2-2y+1=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

a)

\(\left\{{}\begin{matrix}x+y+xy=7\\x^2+y^2+xy=13\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)

Đặt x+y = S, xy = P,ta có hệ

\(\left\{{}\begin{matrix}S+P=17\\S^2-P=13\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}P=S-17\\S^2-S+4=0\end{matrix}\right.\)

\(S^2-S+4>0\)

=> Hệ phương trình vô nghiệm

HPT \(\Leftrightarrow\left\{{}\begin{matrix}3\left(x^2+y^2\right)+2xy+\dfrac{1}{\left(x-y\right)^2}=20\\\left(x-y\right)+\left(x+y\right)+\dfrac{1}{x-y}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)^2+\left(x-y\right)^2+\dfrac{1}{\left(x-y\right)^2}=20\\\left(x-y\right)+\left(x+y\right)+\dfrac{1}{x-y}=5\end{matrix}\right.\)

Đặt \(a=x+y;b=x-y\)

\(\Rightarrow\left\{{}\begin{matrix}2a^2+b^2+\dfrac{1}{b^2}=20\\a+b+\dfrac{1}{b}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a^2+\left(b+\dfrac{1}{b}\right)^2=22\\b+\dfrac{1}{b}=5-a\end{matrix}\right.\)

\(\Rightarrow2a^2+\left(a-5\right)^2=22\)

\(\)Đến đây thì dễ rồi tự làm nhé

a: \(\left\{{}\begin{matrix}\dfrac{-5x+2y}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4\left(-5x+2y\right)+60=3\left(y+27\right)-24x\\7\left(x+1\right)+21y=3\left(6y-5x\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-20x+8y+60=3y+81-24x\\7x+7+21y=18y-15x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-20x+8y-3y+24x=21\\7x+21y-18y+15x=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+5y=21\\22x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x+15y=63\\110x+15y=-35\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-98x=98\\4x+5y=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\5y=21-4x=21+4=25\end{matrix}\right.\)

=>x=-1 và y=5

b: \(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(xy+3x+2y+6\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(xy-2x-2y+4\right)=32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy+3x+2y+6-xy=100\\xy-\left(xy-2x-2y+4\right)=64\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x+2y=94\\2x+2y=60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=34\\2x+2y=60\end{matrix}\right.\)

=>x=34 và y=-4

c: \(\left\{{}\begin{matrix}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{matrix}\right.\)

\(\left\{{}\begin{matrix}xy-x+20y-20=xy\\xy+x-10y-10=xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x+20y=20\\x-10y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10y=30\\x-10y=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x=10y+10=30+10=40\end{matrix}\right.\)

d: ĐKXĐ: \(\left\{{}\begin{matrix}x< >-2y\\x< >-\dfrac{y}{2}\end{matrix}\right.\)

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

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

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

=>\(\left\{{}\begin{matrix}2x+y=1\\2x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=1\\x+2y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=1-2y=1-\dfrac{2}{3}=\dfrac{1}{3}\end{matrix}\right.\)(nhận)

e: ĐKXĐ: x<>-1 và y<>-4

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

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

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

=>\(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\\\dfrac{2}{x+1}+\dfrac{5}{y+4}=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{6}{x+1}+\dfrac{4}{y+4}=-2\\\dfrac{6}{x+1}+\dfrac{15}{y+4}=-21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{11}{y+4}=19\\\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y+4=-\dfrac{11}{19}\\\dfrac{3}{x+1}+2:\dfrac{-11}{19}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{11}{19}-4=-\dfrac{87}{19}\\\dfrac{3}{x+1}=-1-2:\dfrac{-11}{19}=\dfrac{27}{11}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x+1=\dfrac{11}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{9}\\y=-\dfrac{87}{19}\end{matrix}\right.\left(nhận\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}4\left(x^2-x\right)+1+4\left(y^2-2y\right)+4=10\\\left(x^2-x\right)\left(y^2-2y\right)=-\dfrac{3}{2}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2-x=u\\y^2-2y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4u+1+4v+4=10\\uv=-\dfrac{3}{2}\end{matrix}\right.\)

Chắc em tự giải được hệ này, chỉ cần thế là xong

b) Áp dụng bđt Svac-xơ:

\(\dfrac{1}{x}+\dfrac{9}{y}+\dfrac{16}{z}\ge\dfrac{\left(1+3+4\right)^2}{x+y+z}\ge\dfrac{64}{4}=16>9\)

=> hpt vô nghiệm

c) Ở đây x,y,z là các số thực dương

Áp dụng cosi: \(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)=3xyz\)

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