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nhiều bài quá mình chỉ làm được bài 1,3,4,5
bài 2 mình đang suy nghĩ
bạn có thể vào 
để hỏi bài !
\(\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6+3^5\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}=\frac{3^4.\left(3-1\right)}{3^5\left(3+1\right)}\)
\(=\frac{3^4.2}{3^5.4}=\frac{3^4.2}{3^4.3.4}=\frac{2}{12}=\frac{1}{6}\)
P/s: Hoq chắc ạ (: Ms lp 6 lm đại
\(\frac{x}{2}=\frac{y}{3}\)
\(\Leftrightarrow\frac{x}{8}=\frac{y}{12}\)(1)
\(\frac{y}{4}=\frac{z}{5}\)
\(\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)(2)
Từ (1) (2)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
\(\Rightarrow\hept{\begin{cases}x=2.8\\y=2.12\\z=2.15\end{cases}\Rightarrow}\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}\)
Bài 1:
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
\(\Rightarrow2A=2+\frac{3}{2^2}+\frac{4}{2^3}+....+\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=\left(2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)\)
\(\Rightarrow A=\left(2-1\right)+\frac{3}{2^2}+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)-\frac{100}{2^{100}}\)
\(\Rightarrow A=1+\frac{3}{2^2}+\left(\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)
Bài 2:
Giải:
Ta có: \(2n-3⋮n+1\)
\(\Rightarrow\left(2n+2\right)-5⋮n+1\)
\(\Rightarrow2\left(n+1\right)-5⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in\left\{1;-1;5;-5\right\}\)
\(\Rightarrow n\in\left\{0;-2;4;-6\right\}\)
Vậy ...
A = 22+42+62+...+202
= (1.2)2 + (2.2)2 + (3.2)2 + ... + (10.2)2
= 22 .12 + 22.22 + 22.32 + ... + 22 .102
= 22 . (12 + 22 + 32 + ... + 102)
= 4 . 385
= 1540
Đặt A1 = 1/2^1 + 1/2^2 + ... + 1/2^100
A2 = 1/2^2 + 1/2^3 + ... + 1/2^100
A3 = 1/2^3 + 1/2^4 + ... + 1/2^100
....................................
...................................
A100 = 1/2^100
A = 1/2^1 + 2/2^2 + 3/2^3 + 4/2^4 + ... + 100/2^100 =
= (1/2^1+1/2^2 +...+ 1/2^100) + (1/2^2+1/2^3 +...+ 1/2^100) + (1/2^3+1/2^4 +...+ 1/2^100) + ... + (1/2^100) = A1 + A2 + A3 + ... + A100
2^101 A1 = 2^100 + 2^99 + 2^98 + ... + 2 (1)
2^100 A1 = 2^99 + 2^98 + 2^97 + ... + 1 (2)
(2) trừ (1) ---> 2^100 A1 = 2^100 - 1 ---> A1 = (2^100 - 1) / 2^100 = 1 - 1/2^100
Tương tự
2^101 A2 = 2^99 + 2^98 + 2^97 +...+ 2 (3)
2^100 A2 = 2^98 + 2^97 + 2^96 +...+ 1 (4)
(4) trừ (3) ---> 2^100 A2 = 2^99 - 1 ---> A2 = (2^99 - 1) / 2^100 = 1/2 - 1/2^100
Tương tự
A3 = 1/4 - 1/2^100 = 1/2^2 - 1/2^100
A4 = 1/2^3 - 1/2^100
..................................
.................................
A100 = 1/2^99 - 1/2^100
Vậy A = A1 + A2 + A3 +...+ A100 = (1 + 1/2 + 1/2^2 + ... + 1/2^99) - 100/2^100
= 2 A1 - 100/2^100 = 2 - 2/2^100 - 100/2^100 = 2 - 51/2^99
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