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a) Ta có:
\(3x=4y\Rightarrow\frac{x}{4}=\frac{y}{3}\) (1)
\(3y=5z\Rightarrow\frac{y}{5}=\frac{z}{3}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{x}{4}=\frac{y}{3};\frac{y}{5}=\frac{z}{3}.\)
Có: \(\frac{x}{4}=\frac{y}{3}\Rightarrow\frac{x}{20}=\frac{y}{15}.\)
\(\frac{y}{5}=\frac{z}{3}\Rightarrow\frac{y}{15}=\frac{z}{9}.\)
=> \(\frac{x}{20}=\frac{y}{15}=\frac{z}{9}\) và \(x-y-z=1.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{20}=\frac{y}{15}=\frac{z}{9}=\frac{x-y-z}{20-15-9}=\frac{1}{-4}=\frac{-1}{4}.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{20}=-\frac{1}{4}\Rightarrow x=\left(-\frac{1}{4}\right).20=-5\\\frac{y}{15}=-\frac{1}{4}\Rightarrow y=\left(-\frac{1}{4}\right).15=-\frac{15}{4}\\\frac{z}{9}=-\frac{1}{4}\Rightarrow z=\left(-\frac{1}{4}\right).9=-\frac{9}{4}\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(-5;-\frac{15}{4};-\frac{9}{4}\right).\)
Chúc bạn học tốt!
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!
a, Đặt \(\frac{x}{4}=\frac{y}{7}=\frac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=7k\\z=5k\end{matrix}\right.\)
Mà \(yz-xy-z^2=-72\)
\(\Rightarrow35k^2-28k^2-25k^2=-72\\ \Rightarrow k^2\left(35-28-25\right)=-72\\ k^2\cdot\left(-18\right)=-72\\ \Rightarrow k^2=4\\ \Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
Với k = 2
\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot2=8\\y=7\cdot2=14\\z=5\cdot2=10\end{matrix}\right.\)
Với k = -2
\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot\left(-2\right)=-8\\y=7\cdot\left(-2\right)=-14\\z=5\cdot\left(-2\right)=-10\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)\in\left\{\left(8;14;10\right);\left(-8;-14;-10\right)\right\}\)
b, Đặt \(\frac{x}{2}=\frac{y}{7}=\frac{z}{8}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=7k\\z=8k\end{matrix}\right.\)
Mà \(2x^2+xy-xz=54\)
\(\Rightarrow8k^2+14k^2-16k^2=54\\ \Rightarrow k^2\left(8+14-16\right)=54\\ \Rightarrow k^2\cdot6=54\\ \Rightarrow k^2=9\\ \Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)
Với k = 3
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot3=6\\y=7\cdot3=21\\z=8\cdot3=24\end{matrix}\right.\)
Với k = -3
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot\left(-3\right)=-6\\y=7\cdot\left(-3\right)=-21\\z=8\cdot\left(-3\right)=-24\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)\in\left\{\left(6;21;24\right);\left(-6;-21;-24\right)\right\}\)
c, Đặt \(\frac{x+3}{5}=\frac{y-4}{3}=\frac{z-5}{2}=k\Rightarrow\left\{{}\begin{matrix}x=5k-3\\y=3k+4\\z=2k+5\end{matrix}\right.\)
Mà \(2x-3y-z=-26\)
\(\Rightarrow2\left(5k-3\right)-3\left(3k+4\right)-\left(2k+5\right)=-26\\ \Rightarrow10k-6-9k-12-2k-5=-26\\ \Rightarrow-k=-3\\ \Rightarrow k=3\\ \Rightarrow\left\{{}\begin{matrix}x=5\cdot3-3=12\\y=3\cdot3+4=13\\z=2\cdot3+5=11\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(12;13;11\right)\)
a) \(xy=z;yz=4x;xz=9y\Rightarrow xy.yz.xz=z.4x.9y\Rightarrow\left(xyz\right)^2=36xyz\Rightarrow xyz=36\)
Đấy rồi bạn tự thay giá trị vào tìm ra x;y;z
b) Bài này chắc là rút gọn
\(\frac{2x+9}{x+3}+\frac{5x+17}{x+3}-\frac{3x}{x+3}=\frac{2x+9+5x+17-3x}{x+3}=\frac{4x+26}{x+3}=4+\frac{14}{x+3}\)