Bài 1:
Giải:
Ta có: \(\left\{{}\begin{matrix}3x=4y\\5y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=\dfrac{y}{3}\\\dfrac{y}{6}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{6}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)
Mà \(xyz=30\)
\(\Rightarrow240k^3=30\)
\(\Rightarrow k^3=\dfrac{1}{8}\)
\(\Rightarrow k=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=3\\z=2,5\end{matrix}\right.\)
Vậy...
Bài 2: sai đề
Bài 3:
Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\Rightarrow\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)
Ta có: \(x+2y+3z=38\)
\(\Rightarrow2k+1+8k-6+18k+15=38\)
\(\Rightarrow28k=28\)
\(\Rightarrow k=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=1\\z=11\end{matrix}\right.\)
Vậy...
1) Ta có :
\(3x=4y\Rightarrow\dfrac{3x}{12}=\dfrac{4y}{12}\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\) <=> \(\dfrac{x}{8}=\dfrac{y}{6}\)
\(5y=6z\Rightarrow\dfrac{5y}{30}=\dfrac{6z}{30}\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\)
=> \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)
Thay vào đẳng thức xyz = 30
=> 8k.6k.5k = 30
<=> 240k3 = 30
<=> k3 = 8
<=> k = 2
\(\Rightarrow\left\{{}\begin{matrix}x=8.2=16\\y=6.2=12\\z=5.2=10\end{matrix}\right.\)
b) Câu này cũng tương tự câu 1 nha ! Đặt k luôn , còn không bình phương lên rồi dùng tính chất dãy tỉ số bằng nhau .
c) Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\)
=> \(\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)
Thay vào đẳng thức , ta có :
x + 2y + 3z = 2k + 1 + 2(4k - 3) + 3(6k + 5) = 38
=> 28k = 38
=> k = \(\dfrac{19}{14}\)
Vậy .....
