a, \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\&2x-3y+z=6\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{20}\end{matrix}\right.\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
\(\Rightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\&2x-3y+z=6\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=3\\\dfrac{y}{12}=3\\\dfrac{z}{20}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)
Vậy, ...
b, \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{7}\&2x+3y-z=186\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{y}{20}\\\dfrac{y}{20}=\dfrac{z}{28}\end{matrix}\right.\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\&2x+3y-z=186\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=3\\\dfrac{y}{20}=3\\\dfrac{z}{28}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=45\\y=60\\z=84\end{matrix}\right.\)
Vậy, ...
c, Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow x=2k;y=3k;z=5k\)
\(\Rightarrow xyz=2k.3k.5k=1920\Rightarrow30k^3=1920\)
\(\Rightarrow k^3=64\Rightarrow k=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.4=8\\y=3.4=12\\z=5.4=20\end{matrix}\right.\)
Vậy,...
a) x/3 = y/4 ; y/4 = z/5 và 2x - 3y + z = 6
<=> x/3 = y/4 <=> x/12 = y/16 (1)
<=> y/4 = z/5 <=> y/16 = z/20 (2)
Từ (1) và (2) suy ra : x/12 = y/16 = z/20
<=> 2x/24 = 3y/48 = z/20
Áp dụng t/c dãy tỉ số bằng nhau , ta có :
2x/24 = 3y/48 = z/20 = 2x - 3y + z / 24 - 48 + 20 = -6/4 = -3/2
<=> x/3 = -3/2 => x = -9/2
<=> y/4 = -3/2 => y = -6
<=> z/5 = -3/2 => z = -15/2
Vậy x = -9/2 , b = -6 , z = -15/2 .
