bài 1/
a/ \(\dfrac{x-1}{2}\)=\(\dfrac{y-2}{3}\)=\(\dfrac{z-3}{4}\) và 2x+3y-z
b/ \(\dfrac{2x}{3}\)=\(\dfrac{2y}{4}\)=\(\dfrac{4z}{5}\) và x+y+z=49
bài 2/ Tìm các số a\(_1\); a\(_2\);.....; biết:
\(\dfrac{a_1-1}{9}\)=\(\dfrac{a_2-2}{8}\)=.....=\(\dfrac{a_9-9}{1}\) và a\(_1\)+ a\(_2\)+.....+a\(_9\)= 90
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Đọc tiếp
bài 1/
a/ \(\dfrac{x-1}{2}\)=\(\dfrac{y-2}{3}\)=\(\dfrac{z-3}{4}\) và 2x+3y-z
b/ \(\dfrac{2x}{3}\)=\(\dfrac{2y}{4}\)=\(\dfrac{4z}{5}\) và x+y+z=49
bài 2/ Tìm các số a\(_1\); a\(_2\);.....; biết:
\(\dfrac{a_1-1}{9}\)=\(\dfrac{a_2-2}{8}\)=.....=\(\dfrac{a_9-9}{1}\) và a\(_1\)+ a\(_2\)+.....+a\(_9\)= 90
2,
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=...=\dfrac{a_9-9}{1}=\dfrac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}=\dfrac{\left(a_1+a_2+...+a_9\right)-\left(1+2+...+9\right)}{45}=\dfrac{90-45}{45}=\dfrac{45}{45}=1\\ \Rightarrow a_1=a_2=...=a_9=10\)
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:
\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=...=\dfrac{a_9-1}{1}\)
\(=\dfrac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)
\(=\dfrac{\left(a_1+a_2+...+a_9\right)-\left(1+2+...+9\right)}{9+8+...+1}\)
\(=\dfrac{90-45}{45}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a_1-1}{9}=1\Rightarrow a_1-1=9\Rightarrow a_1=10\\\dfrac{a_2-2}{8}=1\Rightarrow a_2-2=8\Rightarrow a_2=10\\\dfrac{a_9-9}{1}=1\Rightarrow a_9-9=1\Rightarrow a_9=10\end{matrix}\right.\)
\(\Rightarrow a_1=a_2=...=a_9=10\)