Tìm x, y, z biết 2x/3=2y/4=4z/5 và x+y+z=49

1) a thiếu đề .

b) \(\dfrac{2x}{3}=\dfrac{2y}{4}=\dfrac{4z}{5}\)

\(\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{2}=\dfrac{z}{\dfrac{5}{4}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{2}=\dfrac{z}{\dfrac{5}{4}}\)

\(=\dfrac{x+y+z}{\dfrac{3}{2}+2+\dfrac{5}{4}}=\dfrac{49}{\dfrac{19}{4}}\)

\(=\dfrac{196}{19}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{196}{19}.\dfrac{3}{2}=\dfrac{294}{19}\\y=\dfrac{196}{19}.2=\dfrac{392}{19}\\z=\dfrac{196}{19}.\dfrac{5}{4}=\dfrac{245}{19}\end{matrix}\right.\)

\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=....=\dfrac{a_9-9}{1}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=...=\dfrac{a_9-1}{1}\)

\(=\dfrac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)

\(=\dfrac{\left(a_1+a_2+...+a_9\right)-\left(1+2+...+9\right)}{9+8+...+1}\)

\(=\dfrac{90-45}{45}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a_1-1}{9}=1\Rightarrow a_1-1=9\Rightarrow a_1=10\\\dfrac{a_2-2}{8}=1\Rightarrow a_2-2=8\Rightarrow a_2=10\\\dfrac{a_9-9}{1}=1\Rightarrow a_9-9=1\Rightarrow a_9=10\end{matrix}\right.\)

\(\Rightarrow a_1=a_2=...=a_9=10\)