Theo bài ra ta có: \(\dfrac{12x-15y}{7}\)=\(\dfrac{20z-12x}{9}\)=\(\dfrac{15y+20z}{11}\)và x+y+z = 48
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{12x-15y}{7}\)=\(\dfrac{20z-12x}{9}\)=\(\dfrac{15y+20z}{11}\)=\(\dfrac{12x-15y+20z-12x+15y-20z}{7+9+11}\)=\(\dfrac{0}{27}\)=0
+ \(\dfrac{12x-15y}{7}\)=0 \(\Rightarrow\)12x =15y
+\(\dfrac{20z-12x}{9}\)=0 \(\Rightarrow\)20z =12x
+\(\dfrac{15y+20z}{11}\)=0 \(\Rightarrow\)20z= 15y
\(\Rightarrow\)12x = 15y=20z
\(\Rightarrow\)\(\dfrac{z}{\dfrac{1}{20}}\)=\(\dfrac{x}{\dfrac{1}{12}}\)=\(\dfrac{y}{\dfrac{1}{15}}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{z}{\dfrac{1}{20}}\)=\(\dfrac{x}{\dfrac{1}{12}}\)=\(\dfrac{y}{\dfrac{1}{15}}\)=\(\dfrac{x+z+y}{\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{15}}\)=\(\dfrac{48}{\dfrac{1}{5}}\)=240
+\(\dfrac{z}{\dfrac{1}{20}}\)=240 \(\Rightarrow\)z= 240 *\(\dfrac{1}{20}\)=12
+\(\dfrac{x}{\dfrac{1}{12}}\)=240 \(\Rightarrow\)x= 240 *\(\dfrac{1}{12}\)=20
+\(\dfrac{y}{\dfrac{1}{15}}\)=240 \(\Rightarrow\)y= 240 * \(\dfrac{1}{15}\)=18
Vậy x= 20; y= 18; z = 12
Ở đây mình nhầm nhé. y = 16 chứ không phải là 18 nha. Bạn sửa lại cho đúng!!!
Áp dụng tính chất dãy tỉ số bằng nhau
\(\dfrac{12x-15y}{7}=\dfrac{20z-12x}{9}=\dfrac{15y-20z}{11}\)
\(=\dfrac{12x-15y+20z-12x+15y-20z}{7+9+11}\)
\(=\dfrac{0}{27}=0\)
\(\Rightarrow12x-15y=0\Rightarrow12x=15y\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{12}\Rightarrow\dfrac{x}{60}=\dfrac{y}{48}\)
\(\Rightarrow\)\(20z-12x=0\Rightarrow20z=12x\)
\(\Rightarrow\dfrac{x}{20}=\dfrac{z}{12}\Rightarrow\dfrac{x}{60}=\dfrac{z}{36}\)
\(\Rightarrow\dfrac{x}{60}=\dfrac{y}{48}=\dfrac{z}{36}=\dfrac{x+y+z}{60+48+36}=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{3}\times60=20;y=\dfrac{1}{3}\times48=16;\)
\(z=\dfrac{1}{3}\times36=12\)
