A=1/2+1/22+1/23+...+1/213+1/214
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VH
0
D
3
1 tháng 8 2017
a) \(1-2+3-4+...+213-214\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(213-214\right)\)(có 107 cặp)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=-107\)
b) \(\left|x-17\right|+13=25\)
\(\Rightarrow\left|x-17\right|=25-13\)
\(\Rightarrow\left|x-17\right|=12\)
\(\Rightarrow\left[{}\begin{matrix}x-17=12\\17-x=12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12+17\\-x=12-17\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=29\\-x=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=29\\x=5\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=29\\x=5\end{matrix}\right.\)
1 tháng 8 2017
\(1-2+3-4+5-6+...+213-214\)
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(213-214\right)\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=\left(-1\right)\left[\left(\dfrac{214-1}{1}+1\right):2\right]\)
\(=-1.107\)
\(=-107\)
\(\left|x-17\right|+13=25\)
\(\Rightarrow\left|x-17\right|=12\)
\(\Rightarrow\left[{}\begin{matrix}x-17=12\\x-17=-12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=29\\x=5\end{matrix}\right.\)
1 tháng 9 2023
a) \(A=1+2+2^2+...+2^{80}\)
\(2A=2+2^2+2^3+...+2^{81}\)
\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)
\(A=2^{81}-1\)
Nên A + 1 là:
\(A+1=2^{81}-1+1=2^{81}\)
b) \(B=1+3+3^2+...+3^{99}\)
\(3B=3+3^2+3^3+...+3^{100}\)
\(3B-B=3+3^2+3^3+...+3^{100}-1-3-3^2-...-3^{99}\)
\(2B=3^{100}-1\)
Nên 2B + 1 là:
\(2B+1=3^{100}-1+1=3^{100}\)
1 tháng 9 2023
2)
a) \(2^x\cdot\left(1+2+2^2+...+2^{2015}\right)+1=2^{2016}\)
Gọi:
\(A=1+2+2^2+...+2^{2015}\)
\(2A=2+2^2+2^3+...+2^{2016}\)
\(A=2^{2016}-1\)
Ta có:
\(2^x\cdot\left(2^{2016}-1\right)+1=2^{2016}\)
\(\Rightarrow2^x\cdot\left(2^{2016}-1\right)=2^{2016}-1\)
\(\Rightarrow2^x=\dfrac{2^{2016}-1}{2^{2016}-1}=1\)
\(\Rightarrow2^x=2^0\)
\(\Rightarrow x=0\)
b) \(8^x-1=1+2+2^2+...+2^{2015}\)
Gọi: \(B=1+2+2^2+...+2^{2015}\)
\(2B=2+2^2+2^3+...+2^{2016}\)
\(B=2^{2016}-1\)
Ta có:
\(8^x-1=2^{2016}-1\)
\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)
\(\Rightarrow2^{3x}-1=2^{2016}-1\)
\(\Rightarrow2^{3x}=2^{2016}\)
\(\Rightarrow3x=2016\)
\(\Rightarrow x=\dfrac{2016}{3}\)
\(\Rightarrow x=672\)
TT
1
NV
11 tháng 3 2023
Là \(\left(\dfrac{1}{2}\right)^2\) hay \(\dfrac{1}{2^2}\) vậy bạn
Những cái sau tương tự
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
5 tháng 4 2023
a, \(\dfrac{7}{22}\) - \(\dfrac{15}{23}\) + \(\dfrac{2022}{2023}\) - \(\dfrac{8}{23}\) + \(\dfrac{15}{22}\)
= ( \(\dfrac{7}{22}\) + \(\dfrac{15}{22}\)) - ( \(\dfrac{15}{23}+\dfrac{18}{23}\)) + \(\dfrac{2022}{2023}\)
= \(\dfrac{22}{22}\) - \(\dfrac{23}{23}\) + \(\dfrac{2022}{2023}\)
= 1 - 1 + \(\dfrac{2022}{2023}\)
= \(\dfrac{2022}{2023}\)
b, - \(\dfrac{2}{11}\) + 5\(\dfrac{5}{6}\) ( 14\(\dfrac{1}{5}\) - 11\(\dfrac{1}{5}\)): 5\(\dfrac{1}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) ( \(\dfrac{71}{5}\) - \(\dfrac{56}{5}\)) : \(\dfrac{11}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) . \(\dfrac{15}{5}\) : \(\dfrac{11}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{2}\) \(\times\) \(\dfrac{2}{11}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{11}\)
= \(\dfrac{33}{11}\)
= 3
c, 2000 + { 20 - [ 4.20220 - (32 + 5):2] }
= 2000 + { 20 - [ 4.1 - (9+5):2]}
= 2000 + { 20 - [ 4 - 14 : 2 ]}
= 2000 + { 20 - [ 4 -7]}
= 2000 + { 20 - (-3)}
= 2000 + 23
= 2023
CA
1
24 tháng 5 2016
a, Số lượng số hạng của A là: (40-21):1+1=20 số (1)
\(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{40}\)
\(=>A>\frac{1}{40}+\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)(20 số hạng)
\(A>\frac{1}{40}\cdot20=\frac{20}{40}=\frac{1}{2}\)
Vậy A> \(\frac{1}{2}\)
b, Từ (1) => \(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{40}\)
=> \(A< \frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\) ( 20 số hạng)
=> A< \(\frac{1}{20}\cdot20=1\)
Vậy A< 1
CD
3
28 tháng 4 2016
Chọn mình nhé 
Ta có:
\(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{40}\)
\(< \frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=1\) (20 p/số 1/20)
Hay A < 1.
Ta lại có:
\(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{40}\)
\(>\frac{1}{40}+\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{2}\) (20 p/số 1/40)
Hay A > 1
Vậy \(\frac{1}{2}< A< 1\)
28 tháng 4 2016
A=1/21+1/22+1/23+...+1/40(có 20 phân số)
A>1/40+1/40+1/40+...+1/40(có 20 phân số)
A>20/40=1/2(1)
A=1/21+1/22+1/23+...+1/40(có 20 phân số)
A<1/20+1/20+1/20+...+1/20(có 20 phân số)
A<20/20=1(2)
Từ (1) và (2)=>1/2<A<1
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{13}}+\frac{1}{2^{14}}\)
\(\Leftrightarrow2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{13}}+\frac{1}{2^{14}}\right)\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}+\frac{1}{2^{13}}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{12}}+\frac{1}{2^{13}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{13}}+\frac{1}{2^{14}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{14}}\)