Tính:
A = 1/2 - 1/2² + 1/2³ - 1/2⁴ + ... + 1/2⁹⁹ - 1/2¹⁰⁰
K
Khách
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0
0
16 tháng 7 2021
a) bằng 9 nha bạn
b) thì mik ko bik làm.
Đúng thì bạn tim giúp mik nha bạn. Thx bạn
TK
6
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
a) \((-2).(-2).(-2) =4.(-2) = -8\)
b) \((-0,5).(-0,5) = 0,25\)
c)
\(\begin{array}{l}\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ = \frac{{1.1.1.1}}{{2.2.2.2}}\\ = \frac{1}{{16}}\end{array}\)
NN
3
NT
27 tháng 8 2021
a. \(\left(2x+1\right)^2-4x\left(x-1\right)=4x^2+4x+1-4x^2+4x=8x+1\)
b. \(\left(x-2\right)\left(x+2\right)-\left(x-1\right)^2=x^2-4-x^2+2x-1=2x-5\)
28 tháng 2 2022
a: =6/36+1/3=1/6+1/3=1/6+2/6=3/6=1/2
b: =3/4-1/2=3/4-2/4=1/4
JK
15 tháng 2 2022
a. \(\dfrac{5}{17}+\dfrac{-5}{34}.\dfrac{2}{5}\)
= \(\dfrac{5}{17}+\dfrac{1}{-17}\)
= \(\dfrac{5}{17}+\dfrac{-1}{17}\)
= \(\dfrac{4}{17}\)
b. \(\dfrac{1}{2}.\dfrac{5}{6}+\dfrac{2}{3}.\dfrac{3}{4}\)
= \(\dfrac{5}{12}+\dfrac{1}{2}\)
= \(\dfrac{5}{12}+\dfrac{6}{12}\)
= \(\dfrac{11}{12}\)
c. \(\left(\dfrac{-2}{5}+\dfrac{1}{3}\right).\left(\dfrac{3}{2}-\dfrac{3}{7}\right)\)
= \(\left(\dfrac{-6}{15}+\dfrac{5}{15}\right).\left(\dfrac{21}{14}-\dfrac{6}{14}\right)\)
= \(\dfrac{-1}{15}.\dfrac{15}{14}\)
= \(\dfrac{-1}{14}\)
d. \(\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{4}\right)\)
= \(\left(\dfrac{2}{2}+\dfrac{1}{2}\right).\left(\dfrac{3}{3}+\dfrac{1}{3}\right).\left(\dfrac{4}{4}+\dfrac{1}{4}\right)\)
= \(\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}\)
= \(\dfrac{5}{2}\)
15 tháng 2 2022
a: \(=\dfrac{5}{34}\cdot\dfrac{2}{5}=\dfrac{2}{34}=\dfrac{1}{17}\)
b: \(=\dfrac{5}{12}+\dfrac{6}{12}=\dfrac{11}{12}\)
c: \(=\dfrac{-6+5}{15}\cdot\dfrac{21-6}{15}=-\dfrac{1}{15}\)
\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
=>\(2A=1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}\)
=>\(2A+A=1-\dfrac{1}{2}+\dfrac{1}{2^2}-...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}+\dfrac{1}{2}-\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
=>\(3A=1-\dfrac{1}{2^{100}}=\dfrac{2^{100}-1}{2^{100}}\)
=>\(A=\dfrac{2^{100}-1}{2^{100}\cdot3}\)