42+(3x+7):2=2mu 5+3 mu 4
(5 mu 4 + 4 mu 7) (8 mu9 - 2mu 7) (2mu 4-4 mu 2)
chung to rang B = 1/2mu 2 cong 1/3 mu 2 cong 1/4 mu 2 cong 1/5 mu 2 cong 1/6 mu 2cong 1/7 mu 2 cong 1/8 mu2 nho hon 1
bai1
5mu3+3mu5
bai2
(x-1)mu3=125
720/(41-(2*x-5))=2mu 3*5
bai3
1 phan 9 * 3 mu 4 * 3 mu n =3 mu 7
(2 mu 2 chia 4)* 2 mu n = 4
bai4
2 * 2mu2 * 2mu3 * 2mu4 *.......*2 mu100
Cach lam bai (× - 4):5=2mu 4 - 3 mu 2
(2×-9):5=7
4(×-3)= 5 mũ 2 - 1 mũ 10
Nhanh nha sap hoc them oi
(x-4):5=24-32
(x-4):5=16-9
(x-4):5=7
x-4=7.5
x-4=35
x=35+4
x=39
(2x-9):5=7
2x-9=7.5
2x-9=35
2x=35:-9
2x=-5
x=-5.2
x=-10
4(x-3)=52-110
4(x-3)=25-1
4(x-3)=24
x-3=24:4
x-3=6
x=6+3
x=9
(2x-9):5=7
2x-9=7x5
2x-9=35
2x=35+9 suy ra 2x=44
x=44:2 suy ra x=22
mình chỉ làm được bài này thôi mong cậu thông cảm (^-^)
Cm 1/2 mu 2 - 1/ 2mu 4 + 1/ 2 mu 6-...-1/2mu 4n -2 -1/2 mu 4n + ...+ 1/ 2 mu 2014 - 1/ 2 mu 2016 < 0,2
Cm 1/2 mu 2 - 1/ 2mu 4 + 1/ 2 mu 6-...-1/2mu 4n -2 -1/2 mu 4n + ...+ 1/ 2 mu 2014 - 1/ 2 mu 2016<0,2
Cm 1/2 mu 2 - 1/ 2mu 4 + 1/ 2 mu 6-...-1/2mu 4n -2 -1/2 mu 4n + ...+ 1/ 2 mu 2014 - 1/ 2 mu 2016<0,2
\(A=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{2014}}-\frac{1}{2^{2016}}\)
\(\Rightarrow2^2A=1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+\frac{1}{2^8}-...+\frac{1}{2^{2012}}-\frac{1}{2^{2014}}\)
\(\Rightarrow2^2A+A=1+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^4}-\frac{1}{2^4}\right)+...+\left(\frac{1}{2^{2014}}-\frac{1}{2^{2014}}\right)-\frac{1}{2^{2016}}\)
\(\Rightarrow5A=1-\frac{1}{2^{2016}}< 1\Rightarrow A< \frac{1}{5}=0,2\)
đây là toán lớp 2 hả?
đây là toán lớp mấy thế
cho s= 2+2 mu 2+2 mu 3+ 2mu 4 +....+2 mu 99 +2 mu 100
a, tinh S
b,CHUNG MINH RANG S CHI HET CHO 3
a, S = 2 + 22 + 23 + 24 + ... + 299 + 2100. 2S = 22 + 23 + 24 + 25 + ... + 2100 + 2101 => 2S - S = S = (22 + 23 + 24 + 25 + ... + 2100 + 2101) - (2 + 22 + 23 + 24 + ... + 299 + 2100) = 2101 - 2. Vậy S = 2101 - 2. b, S = 2 + 22 + 23 + 24 + ... + 299 + 2100 = (2 + 22) + (23 + 24) + ... + (299 + 2100) = 2.(1 + 2) + 23.(1 + 2) + ... + 299.(1 + 2) = (1 + 2).(2 + 23 + ... + 299) = 3.(2 + 23 + ... + 299) => S ⋮ 3. Vậy S ⋮ 3 (đpcm)
A = 2mu 0 + 2 mu 1 + 2mu 2 + ...+ 2 mu 50
A= \(2^0+2^1+2^2+...+2^{50}\)
\(\Rightarrow\)2A =2(\(2^0+2^1+2^2+...+2^{50}\))
\(\Rightarrow\)2A= \(2+2^2+2^3+2^4+...+2^{51}\)
\(\Rightarrow\)2A-A= (\(2+2^2+2^3+2^4+...+2^{51}\))-(\(2+2^2+2^3+2^4+...+2^{50}\))
\(\Rightarrow\)A= \(2^{51}-1\)