4/x=y/21=2/7
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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: ...
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}\right)^2-\left(y+\dfrac{1}{y}\right)^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)\left(x+\dfrac{1}{x}-y-\dfrac{1}{y}\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\x+\dfrac{1}{x}-y-\dfrac{1}{y}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=5\\y+\dfrac{1}{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+1=0\\y^2-2y+1=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
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\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{2}{7}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4\cdot7}{2}=14\\y=\dfrac{21\cdot2}{7}=6\end{matrix}\right.\)
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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 .... :)))
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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) \(\dfrac{x}{7}=\dfrac{-6}{21}\)
\(\Rightarrow\dfrac{x}{7}=\dfrac{-2}{7}\)
\(\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
( 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
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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)
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\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}=\dfrac{28:7}{49:7}=\dfrac{4}{9}\\ Vậy:x=\dfrac{4.9}{4}=9\\ y=\dfrac{4.21}{9}=\dfrac{28}{3}\)
\(\dfrac{x}{2}=\dfrac{3}{y}\\ \Leftrightarrow x.y=2.3=6\\ Vậy:\left[{}\begin{matrix}\left(x;y\right)=\left(1;6\right)=\left(6;1\right)\\\left(x;y\right)=\left(2;3\right)=\left(3;2\right)\end{matrix}\right.\)
Ta có cặp X =14 và Y= 6
=> \(\frac{4}{x}=\frac{4}{14}=\frac{2}{7}\)
=> \(\frac{y}{21}=\frac{6}{21}=\frac{2}{7}\)
=> \(\frac{4}{14}=\frac{6}{21}=\frac{2}{7}\)
y=6,x=14