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\(xy=\dfrac{1}{2};yz=\dfrac{3}{5};xz=\dfrac{27}{10}\)
\(\Rightarrow xy.yz.xz=\dfrac{1}{2}.\dfrac{3}{5}.\dfrac{27}{10}\)
\(\Rightarrow\left(xyz\right)^2=\dfrac{81}{100}\)
\(\Rightarrow\left\{{}\begin{matrix}xyz=\dfrac{9}{10}\\xyz=\dfrac{-9}{10}\\xyz=\dfrac{9}{-10}\\xyz=\dfrac{-9}{-10}\end{matrix}\right.\)
\(xy=\dfrac{1}{2};yz=\dfrac{3}{5};xz=\dfrac{27}{16}\)
\(\Rightarrow xy.yz.xz=\dfrac{1}{2}.\dfrac{3}{5}.\dfrac{27}{16}\)
\(\Rightarrow\left(xyz\right)^2=\dfrac{81}{160}\)
\(\Rightarrow\left\{{}\begin{matrix}xyz=\sqrt{\dfrac{81}{160}}\\xyz=-\sqrt{\dfrac{81}{160}}\end{matrix}\right.\)
\(\frac{x}{3}=\frac{y}{5}\)và x + y = 16
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{16}{8}=2\)
\(\frac{x}{3}=2\Rightarrow x=2.3=6\)
\(\frac{y}{5}=2\Rightarrow y=2.5=10\)
Vậy...
a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{-3}=\dfrac{x.y.z}{5.2.-3}=\dfrac{240}{-30}=-8\)
\(\Rightarrow\dfrac{x}{5}=-8\Rightarrow x=-8.5=-40\)
\(\Rightarrow\dfrac{y}{2}=-8\Rightarrow y=-8.2=-16\)
\(\Rightarrow\dfrac{z}{-3}=-8\Rightarrow z=-8.-3=24\)
Vậy \(x=--40;y=-16\) và \(z=24\)
b) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{2}=\dfrac{x^3-y^3+z^3}{3^3-4^3+2^3}=\dfrac{-29}{-29}=1\)
\(\Rightarrow\dfrac{x}{3}=1\Rightarrow x=3.1=3\)
\(\Rightarrow\dfrac{y}{4}=1\Rightarrow y=1.4=4\)
\(\Rightarrow\dfrac{z}{2}=1\Rightarrow z=1.2=2\)
Vậy \(x=3;y=4\) và \(z=2\)
a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)
b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)
\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)
d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)
\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)
Sửa lại đề theo bạn ns:
Ta có:
\(xy.yz.xz=\dfrac{1}{2}.\dfrac{3}{5}.\dfrac{27}{10}\)
\(\Rightarrow\left(xyz\right)^2=\dfrac{81}{100}\Rightarrow xyz=\pm\dfrac{9}{10}\)
Xét \(xyz=-\dfrac{9}{10}\) ta có:
\(\left\{{}\begin{matrix}x=xyz:yz=-\dfrac{9}{10}:\dfrac{3}{5}=-\dfrac{3}{2}\\y=xyz:xz=-\dfrac{9}{10}:\dfrac{27}{10}=-\dfrac{1}{3}\\z=xyz:xy=-\dfrac{9}{10}:\dfrac{1}{2}=-\dfrac{9}{5}\end{matrix}\right.\).
Xét \(xyz=\dfrac{9}{10}\) ta có:
\(\left\{{}\begin{matrix}x=xyz:yz=\dfrac{9}{10}:\dfrac{3}{5}=\dfrac{3}{2}\\y=xyz:xz=\dfrac{9}{10}:\dfrac{27}{10}=\dfrac{1}{3}\\z=xyz:xy=\dfrac{9}{10}:\dfrac{1}{2}=\dfrac{9}{5}\end{matrix}\right.\)
Vậy............ Chúc bạn học tốt!!!Mạng vs chả lỗi @@!
Ta có:
\(2y=\dfrac{3}{5}\Rightarrow y=\dfrac{3}{10}\)
Thay \(y=\dfrac{3}{10}\) vào \(xy=\dfrac{1}{2}\) ta được:
\(\dfrac{3}{10}x=\dfrac{1}{2}\Rightarrow x=\dfrac{5}{3}\)
Thay \(x=\dfrac{5}{3}\) vào \(xz=\dfrac{27}{10}\) ta được:
\(\dfrac{5}{3}z=\dfrac{27}{10}\Rightarrow z=\dfrac{81}{50}\)
Chúc bạn học tốt!!!