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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{3+13}=\frac{2x}{16}=\frac{25x}{200}=\frac{xy}{200}\)
\(\Rightarrow25x=xy\Rightarrow y=25\)
\(\Rightarrow\frac{x-25}{3}=\frac{x+25}{13}\)
\(\Leftrightarrow13x-325=3x+75\)
\(\Leftrightarrow10x=400\Rightarrow x=40\)
Vậy \(x=40;y=25\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{xy}{2}=\frac{yz}{4,5}=\frac{xz}{8}=\frac{xy+yz+xz}{2+4,5+8}=\frac{29}{14,5}=2\)
\(\Rightarrow xy=4,yz=9,xz=16\)
\(\Rightarrow\left(xy\right).\left(yz\right).\left(xz\right)=4.9.16\)
\(\Rightarrow\left(xyz\right)^2=2^2.3^2.4^2\Rightarrow\left(xyz\right)^2=24^2\Rightarrow\orbr{\begin{cases}xyz=24\\xyz=-24\end{cases}}\)
Nếu xyz = 24 thì \(\hept{\begin{cases}x=\left(xyz\right):\left(yz\right)=24:9=\frac{8}{3}\\y=\left(xyz\right):\left(xz\right)=24:16=\frac{3}{2}\\z=\left(xyz\right):\left(xy\right)=24:4=6\end{cases}}\)
Nếu xyz = -24 thì \(\hept{\begin{cases}x=\left(xyz\right):\left(xz\right)=-24:9=-\frac{8}{3}\\y=-24:16=-\frac{3}{2}\\z=-24:4=-6\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: \(\frac{x^2-yz}{a}=\frac{y^2-xz}{b}=\frac{z^2-xy}{c}\)
\(\Rightarrow\frac{a}{x^2-yz}=\frac{b}{y^2-xz}=\frac{c}{z^2-xy}\Rightarrow\frac{a^2}{\left(x^2-yz\right)^2}=\frac{b^2}{\left(y^2-xz\right)^2}=\frac{c^2}{\left(z^2-xy\right)^2}\) (1)
=> \(\frac{a}{\left(x^2-yz\right)}.\frac{a}{\left(x^2-yz\right)}=\frac{b}{y^2-xz}.\frac{c}{z^2-xy}=\frac{a^2}{\left(x^2-yz\right)^2}=\frac{bc}{\left(y^2-xz\right).\left(z^2-xy\right)}\)
a^2/(x^2-yz)^2 = (a^2-bc)/[(x^2-yz)^2 - (y^2-xz)(z^2-xy)] = (a^2-bc)/[x (x^3 + y^3 + z^3 - 3xyz)] =>
(a^2-bc)/x = [a^2/(x^2 - yz)^2] * (x^3 + y^3 + z^3 - 3xyz) (2)
Thực hiện tương tự ta cũng có
(b^2-ac)/y = [b^2/(y^2 - xz)^2] * (x^3 + y^3 + z^3 - 3xyz) (3)
(c^2-ab)/z = [c^2/(z^2 - xy)^2] * (x^3 + y^3 + z^3 - 3xyz) (4)
Từ (1),(2),(3),(4) => (a^2-bc)/x = (b^2-ac)/y = (c^2-ab)/z.
a) Ta có:
\(\frac{x-y}{3}=\frac{x+y}{13}.\)
\(\Rightarrow\left(x-y\right).13=\left(x+y\right).3\)
\(\Rightarrow13x-13y=3x+3y\)
\(\Rightarrow13x-3x=3y+13y\)
\(\Rightarrow10x=16y\)
\(\Rightarrow x=\frac{16y}{10}\)
\(\Rightarrow x=\frac{16}{10}y\)
\(\Rightarrow x=\frac{8}{5}y.\)
+ Thay \(x=\frac{8}{5}y\) vào vào đề bài ta được:
\(\frac{\frac{8}{5}y-y}{3}=\frac{\frac{8}{5}y+y}{13}=\frac{\frac{8}{5}y.y}{200}\)
\(\Rightarrow\frac{\frac{3}{5}y}{3}=\frac{\frac{13}{5}y}{13}=\frac{\frac{8}{5}y^2}{200}\)
\(\Rightarrow\frac{3}{5}y.\frac{1}{3}=\frac{13}{5}y.\frac{1}{13}=\frac{8}{5}y^2.\frac{1}{200}\)
\(\Rightarrow\frac{1}{5}y=\frac{1}{5}y=\frac{1}{125}y^2\)
\(\Rightarrow\frac{1}{5}y=\frac{1}{125}y^2\)
\(\Rightarrow\frac{1}{5}y-\frac{1}{125}y^2=0\)
\(\Rightarrow\frac{1}{5}y.\left(1-\frac{1}{25}y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{5}y=0\\1-\frac{1}{25}y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\\frac{1}{25}y=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=25\end{matrix}\right.\)
+ TH1: \(y=0.\)
\(\Rightarrow x=\frac{8}{5}.0\)
\(\Rightarrow x=0.\)
+ TH2: \(y=25.\)
\(\Rightarrow x=\frac{8}{5}.25\)
\(\Rightarrow x=40.\)
Vậy \(\left(x;y\right)=\left(0;0\right),\left(40;25\right).\)
Chúc bạn học tốt!
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