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: ...

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

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

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

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

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

\(\Leftrightarrow...\)

\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{2}{7}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4\cdot7}{2}=14\\y=\dfrac{21\cdot2}{7}=6\end{matrix}\right.\)

x = 14

y = 6

Đặt  \(\frac{x}{7}=\frac{y}{4}=a\left(a\ne0\right)\)

\(\Rightarrow\hept{\begin{cases}x=7a\\y=4a\end{cases}}\)

Mà  \(x-y=21\)

\(\Leftrightarrow7a-4a=21\)

\(\Leftrightarrow3a=21\)

\(\Leftrightarrow a=7\)

\(\Rightarrow\hept{\begin{cases}x=7a=49\\y=4a=28\end{cases}}\)

Vậy .... :)))

X=49, Y=28

x = 49 

y = 28 

ủng hộ mk nha

a) \(\dfrac{x}{2}=\dfrac{3}{y}=\dfrac{2}{4}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{3}{y}=\dfrac{1}{2}\)

Ta thấy mẫu số của phân số thứ nhất bằng mẫu số của phân số thứ ba nên tử số của phân số thứ nhất cũng chính là tử số của phân số thứ ba.

\(\Rightarrow x=1\)

Ta có:

\(\dfrac{1}{2}=\dfrac{3}{y}\)

\(\Rightarrow\dfrac{3}{6}=\dfrac{3}{y}\)

\(\Rightarrow y=6\)

Vậy \(x=1;y=6\)

b) \(\dfrac{x}{7}=\dfrac{-6}{21}\)

\(\Rightarrow\dfrac{x}{7}=\dfrac{-2}{7}\)

\(\Rightarrow x=2\)

( 2 x y + 2/15 ) x 3 = 4/5

( 2 x y + 2/15 )      = 4/5 : 3 

( 2 x y + 2/15 )      =   4/15

 2 x y                    = 4/15 - 2/15 

2 x y                     =     2/15

     y                      =     2/15 :2 

   y                          =    1/15

(2 x y + 2/15) x 3 = 4/5 

2 x y + 2/15) = 4/5 : 3 

2 x y + 2/15 = 4/15 

2 x y = 4/15 - 2/15 

2 x y = 2/15 

y = 2/15 : 2 

y = 1/15 

7/9 x (2 - 1/3 x y) = 14/15 

(2 - 1/3 x y) = 14/15 : 7/9 

(2 - 1/3 x y) = 6/5 

2 - y = 6/5 x 1/3 

2 - y = 2/5 

y = 2/5 + 2 

y = 12/5 

4/21 + 5 x y - 8/7 = 1/3 

4/21 + 5 x y = 1/3 + 8/7 

4/21 + 5 x y = 31/21 

5 x y = 31/21 - 4/21 

5 x y = 9/7 

y = 9/7 : 5 

y = 9/35 

7/12 x y - 3/12 x y = 5 

y x (7/12 - 3/12) = 5 

y x 1/3 = 5 

y = 5 : 1/3 

y = 15 

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

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

Có hệ \(\left\{{}\begin{matrix}u+v=7\\u^2-v^2=21\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}u+v=7\\\left(u+v\right)\left(u+v\right)=21\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u+v=7\\u-v=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=5\\v=2\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=5\\xy=2\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}\left(x+y\right)^2=5+2.2=9\\xy=2\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x+y=3\\x+y=-3\end{matrix}\right.\\xy=2\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=3-y\\\left(3-y\right)y=2\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=3-y\\y^2-3y+2=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=3-y\\\left(y-2\right)\left(y-1\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=3-y\\\left[{}\begin{matrix}y=2\\y=1\end{matrix}\right.\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=-3\\xy=2\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=-2-y\\\left(-2-y\right)y=2\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=-2-y\\y^2+2y+2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=-2-y\\\left(y+1\right)^2+1=0\end{matrix}\right.\)(vô nghiệm)

Vậy hpt có hai nghiệm duy nhất (1,2),(2,1)

\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}=\dfrac{28:7}{49:7}=\dfrac{4}{9}\\ Vậy:x=\dfrac{4.9}{4}=9\\ y=\dfrac{4.21}{9}=\dfrac{28}{3}\)

\(\dfrac{x}{2}=\dfrac{3}{y}\\ \Leftrightarrow x.y=2.3=6\\ Vậy:\left[{}\begin{matrix}\left(x;y\right)=\left(1;6\right)=\left(6;1\right)\\\left(x;y\right)=\left(2;3\right)=\left(3;2\right)\end{matrix}\right.\)

\(3,\\ \dfrac{42}{54}=\dfrac{42:6}{54:6}=\dfrac{7}{9}\\ \dfrac{42}{54}=\dfrac{7}{x}=\dfrac{7}{9}\\ Vậy:x=\dfrac{7.9}{7}=9\)