Câu hỏi :
A) 1 + 1= ...
B) 2 + 2 = ....
AI NHANH NHẤT MK TK CHO
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Ta có:
3a + 1 = (b + 1)2
=> 3a + 1 = (b + 1).(b + 1)
=> 3a + 1 = (b + 1).b + (b +1)
=> 3a + 1 = b2 + b + b + 1
=> 3a = b2 + 2b
=> 3a = b.(b + 2)
=> 3a = 31 = b.(b + 2) = 1.3 => a = b = 1
Vì với a nguyên dương > 1 thì 3a khi phân tích thành tích 2 số bất kì thì khoảng cách giữa 2 số đó luôn > 2
Vậy a = b = 1
Mình là Nguyên đây! Nhưng làm sao bn chắc chắn rằng3^a=3^1 dc chứ???
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(A=\frac{1-\frac{1}{3^{100}}}{2}\)
\(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(3B=\frac{5.3}{4.7}+\frac{5.3}{7.10}+\frac{5.3}{10.13}+...+\frac{5.3}{25.28}\)
\(3B=5\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)
\(3B=5\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(3B=5\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(3B=5\cdot\frac{3}{14}=\frac{15}{14}\)
\(B=\frac{15}{14}:3=\frac{5}{14}\)
a) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}\)
b) \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{5}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{5}{3}.\left(\frac{1}{10}-\frac{1}{13}\right)+...+\frac{5}{3}.\left(\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\frac{3}{14}\)
\(\Rightarrow B=\frac{5}{14}\)
B = 1+ 2 + 3 + ... + 98 + 99
số số hạng từ 1 đến 99 là : (99 - 1) : 1 + 1 = 99
=) B = (99+1) . 99 : 2 = 4950
vậy B = 4950
\(2010^2-2009^2+2008^2-...+2^2-1^2\)
\(=-\left(1^2-2^2+3^2-...+2009^2-2010^2\right)\)
\(=-\left[1^2+2^2+...+2009^2+2010^2-\left(2^2+4^2+...+2010^2\right)\right]\)
\(=-\left[\frac{2010.\left(2010-1\right)\left(2.2010-1\right)}{6}-2^2\left(1^2+2^2+...+1005^2\right)\right]\)
\(=-\left[2704847285-2^2.\frac{1005\left(1005-1\right)\left(2.1005-1\right)}{6}\right]\)
\(=-\left(2704847285-1351414120\right)=1353433165\)
a,1+1=2
b,2+2=4
tk mk nha bn
A) 1+1=2
B) 2+2=4