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Áp dụng tính chất dãy tỉ số bằng nhau
\(\dfrac{a1-1}{9}=\dfrac{a2-2}{8}=\dfrac{a3-3}{7}=...=\dfrac{a9-9}{1}=\dfrac{a1-1+a2-2+a3-3+...+a9-9}{9+8+7+...+1}=\dfrac{\left(a1+a2+...+a9\right)-\left(1+2+...+9\right)}{9+8+7+...+1}=\dfrac{\left(a1+a2+...+a9\right)-\left[9.\left(9+1\right):2\right]}{45}=\dfrac{90-45}{45}=\dfrac{45}{45}=1\)\(\Rightarrow\dfrac{a1-1}{9}=1\Rightarrow a1-1=9\Rightarrow a1=9+1\Rightarrow a1=10\)
\(\dfrac{a2-2}{8}=1\Rightarrow a2-2=8\Rightarrow a2=8+2\Rightarrow a2=10\)
\(\dfrac{a3-3}{7}=1\Rightarrow a3-3=7\Rightarrow a3=7+3\Rightarrow a3=10\)
\(...\)
\(\dfrac{a9-9}{1}=1\Rightarrow a9-9=1\Rightarrow a9=1+9\Rightarrow a9=10\)
Vậy a1 = a2 = a3 = ... = a9
Áp dụng tính chất dãy tỉ số bằng nhau ta có :0
\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=..............=\dfrac{a_9-9}{1}=\dfrac{\left(a_1+a_2+......+a_9\right)-\left(1+2+....+9\right)}{9+8+..+1}\)
\(=\dfrac{90-45}{45}=1\)
+) \(\dfrac{a_1-1}{9}=1\Leftrightarrow a_1=10\)
+) \(\dfrac{a_2-1}{8}=1\Leftrightarrow a_2=10\)
........................
+) \(\dfrac{a_9-9}{1}=1\Leftrightarrow a_9=10\)
Vậy \(a_1=a_2=..........=a_9=10\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=\dfrac{a_3-3}{7}=...=\dfrac{a_9-9}{1}\)
\(=\dfrac{a_1+a_2+...+a_9-\left(1+2+...+9\right)}{9+8+7+...+1}\)\(=\dfrac{90-45}{45}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a_1-1}{9}=1\\\dfrac{a_2-2}{8}=1\\.................\\\dfrac{a_9-9}{1}=1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a_1-1=9\\a_2-2=8\\.................\\a_9-9=1\end{matrix}\right.\)\(\Rightarrow a_1=a_2=...=a_9=10\)
\(M=\left(\dfrac{1252}{417.762}-\dfrac{4}{417.762}\right)-\left(\dfrac{1}{139}+\dfrac{5}{139}\right).\dfrac{3809}{762}\)
\(M=\left(\dfrac{1252-4}{317754}\right)-\dfrac{6}{139}.\dfrac{3809}{762}\)
\(M=\left(\dfrac{1248}{317754}\right)-\dfrac{22854}{105918}\)
\(M=\dfrac{208}{52959}-\dfrac{3809}{17653}\)
\(M=\dfrac{3671824}{934885227}-\dfrac{201720831}{934885227}\)
\(M=\dfrac{-198049007}{934885227}\)
Lời giải:
Đặt $\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_n}{a_{n+1}}=t$
Áp dụng TCDTSBN:
$t=\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+a_3+....+a_n}{a_2+a_3+....+a_{n+1}}$
$\Rightarrow t^n=\left[\frac{a_1+a_2+a_3+....+a_n}{a_2+a_3+....+a_{n+1}}\right]^n(*)$
Lại có:
$\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}....\frac{a_n}{a_{n+1}}=t.t.t....t$
$\Rightarrow \frac{a_1}{a_{n+1}}=t^n(**)$
Từ $(*)$ và $(**)$ ta có:
$\left[\frac{a_1+a_2+a_3+....+a_n}{a_2+a_3+....+a_{n+1}}\right]^n=\frac{a_1}{a_{n+1}}$ (đpcm)
\(a_1< a_2< a_3\\ \Rightarrow a_1+a_2+a_3=a_3+a_3+a_3=3a_3\\ a_4< a_5< a_6\\ \Rightarrow a_4+a_5+a_6=a_6+a_6+a_6=3a_6\\ a_7< a_8< a_9\\ \Rightarrow a_7+a_8+a_9=a_9+a_9+a_9=3a_9\\ \dfrac{a_1+a_2+...+a_9}{a_3+a_6+a_9}< \dfrac{3a_3+3a_6+3a_9}{a_3+a_6+a_9}=\dfrac{3\left(a_3+a_6+a_9\right)}{a_3+a_6+a_9}=3\left(ĐPCM\right)\)
\(A=3\dfrac{1}{417}.\dfrac{1}{762}-\dfrac{1}{139}+\dfrac{761}{762}-\dfrac{4}{417.762}+\dfrac{5}{139}\)
\(=\left(\dfrac{1252}{417.762}-\dfrac{4}{417.762}\right)+\left(-\dfrac{1}{139}+\dfrac{5}{139}\right)+\dfrac{761}{762}\)
\(=\dfrac{1248}{417.762}+\dfrac{4}{139}+\dfrac{761}{762}=\dfrac{1248}{417.672}+\dfrac{12.762}{417.762}+\dfrac{761.417}{417.762}\)
\(=\dfrac{327729}{317754}\)
Bài 2:
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=...=\dfrac{a_9-9}{1}=\dfrac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}=\dfrac{\left(a_1+a_2+...+a_9\right)-\left(1+2+...+9\right)}{9+8+...+1}\)
\(=\dfrac{90-45}{45}=1\)
+) \(\dfrac{a_1-1}{9}=1\Rightarrow a_1=10\)
+) \(\dfrac{a_2-2}{8}=1\Rightarrow a_2=10\)
...
+) \(\dfrac{a_9-9}{1}=1\Rightarrow a_9=10\)
Vậy \(a_1=a_2=...=a_9=10\)
giải bài 1 đê , đừng có lấy máy tính ra tính nhen