\(M=\left(\dfrac{1252}{417.762}-\dfrac{4}{417.762}\right)-\left(\dfrac{1}{139}+\dfrac{5}{139}\right).\dfrac{3809}{762}\)

\(M=\left(\dfrac{1252-4}{317754}\right)-\dfrac{6}{139}.\dfrac{3809}{762}\)

\(M=\left(\dfrac{1248}{317754}\right)-\dfrac{22854}{105918}\)

\(M=\dfrac{208}{52959}-\dfrac{3809}{17653}\)

\(M=\dfrac{3671824}{934885227}-\dfrac{201720831}{934885227}\)

\(M=\dfrac{-198049007}{934885227}\)

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

\(A=3\dfrac{1}{417}.\dfrac{1}{762}-\dfrac{1}{139}+\dfrac{761}{762}-\dfrac{4}{417.762}+\dfrac{5}{139}\)

\(=\left(\dfrac{1252}{417.762}-\dfrac{4}{417.762}\right)+\left(-\dfrac{1}{139}+\dfrac{5}{139}\right)+\dfrac{761}{762}\)

\(=\dfrac{1248}{417.762}+\dfrac{4}{139}+\dfrac{761}{762}=\dfrac{1248}{417.672}+\dfrac{12.762}{417.762}+\dfrac{761.417}{417.762}\)

\(=\dfrac{327729}{317754}\)

Câu 2

(a+3)(b-4)-(a-3)(b+4)=0

=>ab-4a+3b-12-ab-4a+3b+12=0

=>-8a=-6b

=>a/b=3/4

=>a/3=b/4

Đặt 761=a; 139=b

\(M=\left(3+\dfrac{1}{3b}\right)\cdot\dfrac{1}{a+1}-\dfrac{1}{b}\cdot\left(4+\dfrac{a}{a+1}\right)-\dfrac{4}{\left(a+1\right)\cdot3b}+\dfrac{5}{b}\)

\(=\dfrac{9b+1}{3b}\cdot\dfrac{1}{a+1}-\dfrac{1}{b}\cdot\dfrac{4a+4}{a+1}-\dfrac{4}{3b\left(a+1\right)}+\dfrac{5}{b}\)

\(=\dfrac{9b+1-3\left(4a+4\right)-4+15\left(a+1\right)}{3b\left(a+1\right)}\)

\(=\dfrac{9b+1-12a-12-4+15a+15}{3b\left(a+1\right)}\)

\(=\dfrac{3a+9b}{3b\left(a+1\right)}=\dfrac{3\left(a+3b\right)}{3b\left(a+1\right)}=\dfrac{a+3b}{b\left(a+1\right)}\)

\(=\dfrac{761+3\cdot139}{139\left(761+1\right)}=\dfrac{589}{52959}\)

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:

\(\dfrac{a_1-1}{9}=\dfrac{a_2-2}{8}=\dfrac{a_3-3}{7}=...=\dfrac{a_9-9}{1}\)

\(=\dfrac{a_1+a_2+...+a_9-\left(1+2+...+9\right)}{9+8+7+...+1}\)\(=\dfrac{90-45}{45}=1\)

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

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)

các bn cho mk xin lỗi đây là toán lớp 7 nha

\(\dfrac{a_1-1}{100}=\dfrac{a_2-2}{99}=\dfrac{a_3-3}{98}=....=\dfrac{a_{100}-100}{1}=\dfrac{a_1-1+a_2-2+a_3-3+...+a_{100}-100}{100+99+98+...+1}=\dfrac{\left(a_1+a_2+a_3+....+a_{100}\right)-\left(1+2+3+...+100\right)}{100+99+98+....+1}=\dfrac{10100-5050}{5050}=\dfrac{5050}{5050}=1\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a_1-1}{100}=1\Leftrightarrow a_1=1.100+1=101\\\dfrac{a_2-2}{99}=1\Leftrightarrow a_2=1.99+2=101\\..........................................\\\dfrac{a_{100}-100}{1}=1\Leftrightarrow a_{100}=1.1+100=101\end{matrix}\right.\Leftrightarrow a_1=a_2=a_3=...=a_{100}=101\)

I LOVE YOU