\(\frac{8}{x}\) = \(\frac{y}{21}\) = \(\frac{-98}{y}\)
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\(\frac{x}{y}=\frac{3}{2}\Rightarrow\frac{x}{3}=\frac{y}{2}\left(1\right)\)
\(\frac{y}{z}=\frac{2}{-8}\Rightarrow\frac{y}{2}=\frac{z}{-8}\left(2\right)\)
\(\text{Từ}\left(1\right)\left(2\right)\Rightarrow\frac{x}{3}=\frac{y}{2}=\frac{z}{-8}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{3}=\frac{y}{2}=\frac{z}{-8}=\frac{3y-x+\frac{1}{2}z}{2\cdot3-3+\frac{1}{2}\cdot\left(-8\right)}=\frac{21}{7}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{3}=3\\\frac{y}{2}=3\\\frac{z}{-8}=3\end{cases}\Rightarrow\hept{\begin{cases}x=9\\y=6\\z=-24\end{cases}}}\)
Vậy x = 9; y = 6; z = -24
\(\frac{x}{y}=\frac{3}{2}\Rightarrow\frac{x}{3}=\frac{y}{2}\)
Và \(\frac{y}{z}=\frac{2}{-8}\Rightarrow\frac{y}{2}=\frac{z}{-8}\)
\(\frac{x}{3}=\frac{y}{2}=\frac{z}{-8}=\frac{3y}{6}=\frac{\frac{1}{2}z}{-4}=\frac{3y-x+\frac{1}{2}z}{6-3+\left(-4\right)}=\frac{21}{-1}=-21\)
\(\Rightarrow\hept{\begin{cases}x=-63\\y=-42\\z=168\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge16\\y\ge9\end{matrix}\right.\)
Từ pt thứ nhất của hệ:
\(\frac{8xy}{x^2+y^2+6xy}+\frac{17}{8}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{21}{4}\)
\(\Leftrightarrow\frac{8}{\frac{x}{y}+\frac{y}{x}+6}+\frac{17}{8}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{21}{4}\)
Đặt \(\frac{x}{y}+\frac{y}{x}=t\ge2\)
\(\Rightarrow\frac{8}{6+t}+\frac{17}{8}t=\frac{21}{4}\)
\(\Leftrightarrow\frac{17}{8}t^2+\frac{15}{2}t-\frac{47}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-\frac{94}{17}< 0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\frac{x}{y}+\frac{y}{x}=2\Leftrightarrow x^2+y^2=2xy\)
\(\Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\)
Thay xuống pt dưới:
\(\sqrt{x-16}+\sqrt{x-9}=7\)
\(\Leftrightarrow\sqrt{x-16}-3+\sqrt{x-9}-4=0\)
\(\Leftrightarrow\frac{x-25}{\sqrt{x-16}+3}+\frac{x-25}{\sqrt{x-9}+4}=0\)
\(\Leftrightarrow...\)
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a./ \(\frac{x}{5}=\frac{y}{7}=\frac{z}{4}=\frac{x-y+z}{5-7+4}=\frac{-10}{2}=-5\)
\(\Rightarrow x=-25;y=-35;z=-20\)
b./ \(\frac{x}{5}=\frac{y}{-4}=\frac{z}{-7}=\frac{x+y-z}{5-4-\left(-7\right)}=\frac{-40}{6}=-5\)
\(\Rightarrow x=-25;y=20;z=35\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a; \(\dfrac{-x}{4}\) = \(\dfrac{-2}{x}\)
-\(x.x\) = -2.4
-\(x^2\) = -8
\(x^2\) = 8
\(\left[{}\begin{matrix}x=-\sqrt{8}\\x=\sqrt{8}\end{matrix}\right.\)
Vậy \(x\in\) {-\(\sqrt{8}\); \(\sqrt{8}\)}
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
x+y+y+z+z+x=\(\frac{13}{12}\)
2(x+y+z)=\(\frac{13}{12}\)
=>x+y+z=\(\frac{13}{24}\)
z=(x+y+z)-(x+y)
y=y+z-z
x=x+Y-y
đề bài là CM hay là tính x;y ??
Ác Quỷ : tính xy ạ