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V
0
2 tháng 9 2020
\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\)
\(2A-A=A\)
\(=1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(=1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{49}}-\frac{1}{2^{50}}\)
\(=1-\frac{1}{2^{50}}< 1\)
\(\Rightarrow A< 1\)
2 tháng 9 2020
\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=\text{}\text{}1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}\)
Vậy \(A\)< 1
11 tháng 8 2018
\(\frac{1}{7}\times\frac{1}{49}\times49^2=\frac{1}{7}\times\frac{49^2}{49}=\frac{1}{7}\times49=\frac{49}{7}=\frac{7\times7}{7}=7\)
23 tháng 10 2020
Ta có: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\forall n\ge1,n\inℤ\)
Áp dụng:
\(A=1^3+2^3+3^3+...+49^3=\left(1+2+3+...+49\right)^2=\left(\frac{49.50}{2}\right)^2=1500625\)
NP
3
21 tháng 9 2018
(5x + 1)2 = 36/49
=> (5x + 1)2 = (6/7)2
=> \(\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{cases}}\)
21 tháng 9 2018
Làm từ phần b nha
b) \(\left(x-\frac{1}{9}\right)^3=\frac{2}{3}^6\)
\(\Rightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{1}{3}\right)^6\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1^6}{3^6}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1}{3^6}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1}{729}\)
\(\Rightarrow x-\frac{2}{9}=\frac{1}{9}\)
\(x=\frac{1}{9}+\frac{2}{9}\)
\(x=\frac{3}{9}=\frac{1}{3}\)
c) Sai đề rồi, xem lại đi
d) \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4< 0\)
\(\Rightarrow\frac{10000y^4-4000y^3+600y^3-40y+10000x^2+122501-70000x}{10000}< 0\)
=> Sai \(\forall y\inℝ\)
25 tháng 7 2016
(5x+1)2 =\(\frac{36}{49}\)
(5x+1)2=\(\frac{6}{7}^2\)
=>(5x+1)=\(\frac{6}{7}\)
5x =\(\frac{6}{7}\)-1=\(\frac{-1}{7}\)
x = \(\frac{-1}{7}\) :5
x= \(\frac{-1}{35}\)
15 tháng 8 2017
a, \(A=1+2+2^2+2^3+..........+2^{49}+2^{50}\)
\(\Leftrightarrow2A=2+2^2+..............+2^{50}+2^{51}\)
\(\Leftrightarrow2A-A=\left(2+2^2+.........+2^{51}\right)-\left(1+2+......+2^{50}\right)\)
\(\Leftrightarrow A=2^{51}-1\)
15 tháng 8 2017
a) \(A=1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\)
\(\Rightarrow2A=2\left(1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\right)\)
\(2A=2+2^2+2^3+2^4+2^5+...+2^{50}+2^{51}\)
\(\Rightarrow2A-A=A=\left(2+2^2+2^3+2^4+2^5+...+2^{50}+2^{51}\right)-\left(1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\right)\)
\(A=2^{51}-1\) vậy \(A=2^{51}-1\)
b) \(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+\left(\dfrac{1}{2}\right)^5+...+\left(\dfrac{1}{2}\right)^{99}+\left(\dfrac{1}{2}\right)^{100}\)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\)
\(\Rightarrow2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\right)\)
\(2B=\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+\dfrac{2}{2^4}+\dfrac{2}{2^5}+...+\dfrac{2}{2^{99}}+\dfrac{2}{2^{100}}\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(\Rightarrow2B-B=B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\right)\)
\(B=1-\dfrac{1}{2^{100}}\) vậy \(B=1-\dfrac{1}{2^{100}}\)
TA
2
17 tháng 10 2015
S = 1 + 3 + 32 + ... + 3100
3S = 3 + 32 + ... + 3101
3S - S = 3101 - 1
2S = 3101 - 1
S = \(\frac{3^{101}-1}{2}\)
B = 1 + 5 + 52 + ... + 549
5B = 5 + 52 + ... + 550
5B - B = 550 - 1
4B = 550 - 1
B = \(\frac{5^{50}-1}{4}\)