Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Theo bài ra ta cs
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\left(1\right)\)
\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\left(2\right)\)
Từ (1) ; (2) => \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)
ADTC dãy tỉ số bằng nhau ta cs
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{8}=2\\\frac{y}{12}=2\\\frac{z}{15}=2\end{cases}\Rightarrow\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}}\)
Như vậy ta chọn : A
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) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)
\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)
Ta có:\(\frac{x}{3}=\frac{y}{2};\frac{y}{4}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{10}=\frac{30}{20}=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}.6=9\)
\(\Rightarrow y=\frac{3}{2}.4=6\)
\(\Rightarrow z=\frac{3}{2}.10=15\)
Giải:
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}=\frac{\left(x-1\right)+\left(y-2\right)+\left(z-3\right)}{3+4+5}=\frac{\left(x+y+z\right)-6}{12}=\frac{30-6}{12}=2\)
=> \(\hept{\begin{cases}\frac{x-1}{3}=2\\\frac{y-2}{4}=2\\\frac{z-3}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x-1=6\\y-2=8\\z-3=15\end{cases}}\) => \(\hept{\begin{cases}x=7\\y=10\\z=18\end{cases}}\)
Vậy ...
\(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}=\frac{x-1+y-2+z-3}{3+4+5}=\frac{30-6}{12}=\frac{24}{12}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{x-1}{3}=2\\\frac{y-2}{4}=2\\\frac{z-3}{5}=2\end{cases}\Rightarrow}\hept{\begin{cases}x-1=6\\y-2=8\\z-3=10\end{cases}\Rightarrow}\hept{\begin{cases}x=7\\y=10\\z=13\end{cases}}\)
2x=3y
=>\(\dfrac{x}{3}=\dfrac{y}{2}\)
=>\(\dfrac{x}{6}=\dfrac{y}{4}\)
\(5y=4z\)
=>\(\dfrac{y}{4}=\dfrac{z}{5}\)
=>\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{5}\)
mà x+y+z=-30
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y+z}{6+4+5}=\dfrac{-30}{15}=-2\)
=>\(x=-2\cdot6=-12;y=-2\cdot4=-8;z=-2\cdot5=-10\)