Ta có x/8=y/12=z/15
Theo t/c của dãy tỉ số bằng nhau , ta có
x/8=y/12=z/15=x+y-z/8+12-15=10/5=2
x/2=2 , x=4
Ta cóx/2=y/3;y/4=z/5
=>x/8=y/12=z/15
Áp dụng t/c của dãy tỉ số bằng nhau ta có:
x/8=y/12=z/15=x+y-z/8+12-15=10/5=2
=> x=16,y=24,z=30
\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}=>\frac{x}{8}=\frac{y}{12};\frac{y}{12}=\frac{z}{15}\)
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
Nên x = 2.8 = 16
y = 2.12 = 24
z= 2. 15 = 30
\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\)
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{1}{4}.\frac{x}{2}=\frac{1}{4}.\frac{y}{3}=\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{1}{3}.\frac{y}{4}=\frac{1}{3}.\frac{z}{5}=\frac{y}{12}=\frac{z}{15}\)
Áp dụng tính chất DTS bằng nhau:
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\Rightarrow\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
\(\frac{x}{8}=2\Rightarrow x=2.8=16\)
\(\frac{y}{12}=2\Rightarrow y=2.12=24\)
\(\frac{z}{15}=2\Rightarrow z=2.15=30\)
Vậy...
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\)(1)
\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)
Đặt \(\frac{x}{8}=\frac{y}{12}=\frac{z}{18}\)= k => x = 8k ; y = 12k ; z = 15k
x + y - z = 10 => 8k + 12k - 15k = 10 => 5k = 10 => k = 2
Do đó :
\(\frac{x}{8}=2\Rightarrow x=2.8=16\)
\(\frac{y}{12}=2\Rightarrow y=2.12=24\)
\(\frac{z}{15}=2\Rightarrow z=2.15=30\)
Vậy ...
x/8=y/12=z/15=x+y-z/8+12+15=10/5=2
x=16,y=24,z=30
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}\)=2
\(\frac{x}{8}\)= 2 => x = 2.8 = 16
\(\frac{y}{12}\)= 2 => y = 12.2 = 24
\(\frac{z}{15}\)= 2 => z = 15.2 = 30
Vậy x = 16, y = 24, z = 30