\(\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z-3}{5}\)
Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z-3}{5}\)
\(=\dfrac{x-1+y-2+z-3}{3+4+5}\)
\(=\dfrac{x+y+z-6}{12}\)
\(=\dfrac{30-6}{12}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{3}=2\\\dfrac{y-2}{4}=2\\\dfrac{z-3}{5}=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=6\Rightarrow x=7\\y-2=8\Rightarrow y=10\\z-3=10\Rightarrow z=13\end{matrix}\right.\)
Có : \(\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z-3}{5}\) =k và x+y+z=30
=>x-1=3k ; y-2=4k ; z-3 = 5k
=> (x-1)+(y-2)+(z-3)=30-1-2-3
=>(x-1)+(y-2)+(z-3) = 24
=> 3k +4k + 5k = 24
=> k(3+4+5)=24
=>k=24:12=2
*x-1=3k => x=3k+1=3.2+1=7
*y-2=4k=>y=4k+2=4.2+2=10
*z-3=5k=>z=5k+3=5.2+3=13
Vậy : x=7;y=10;z=13
Can you tick for me!
Theo bài ra ta có : \(\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z-3}{5}\)
\(x+y+z=30\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z-3}{5}=\dfrac{\left(x-1\right)+\left(y-2\right)+\left(z-3\right)}{3+4+5}\\ =\dfrac{x-1+y-2+z-3}{12}\\ =\dfrac{\left(x+y+z\right)-\left(1+2+3\right)}{12}\\ =\dfrac{30-6}{12}\\ =\dfrac{24}{12}\\ =2\\ \)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{3}=2\Rightarrow x-1=6\Rightarrow x=7\\\dfrac{y-2}{4}=2\Rightarrow y-2=8\Rightarrow y=10\\\dfrac{z-3}{5}=2\Rightarrow z-3=10\Rightarrow z=13\end{matrix}\right.\\ \)
\(\text{Vậy }x=7\\ y=10\\ z=13\)
Theo tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z-3}{5}=\dfrac{x-1+y-2+z-3}{3+4+5}=\dfrac{\left(x+y+z\right)+\left(-1-2-3\right)}{12}=\dfrac{30-6}{12}=\dfrac{24}{12}=2\)
\(\Rightarrow\dfrac{x-1}{3}=2\Rightarrow x-1=6\Rightarrow x=7\)
\(\dfrac{y-2}{4}=2\Rightarrow y-2=8\Rightarrow y=10\)
\(\dfrac{z-3}{5}=2\Rightarrow z-3=10\Rightarrow z=13\)
Vậy x=7,y=10,z=13
