A)9mũ2: 3mũ3=
b) 5mũ2 .25mũ 2=
c)1/3mũ2 . 1/9.3mũ2=
Cho mk kết quả nhanh nha
K
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 7 2023
a, 21.52.17 = 2.25.17 = 50.17 = 850
b, 22 + 23 + 24 = 4 + 8 + 16 = 28
c, 25.3 + 24:8 + 50: 52
= 32.3 + 16:8 + 50:25
=96 + 2 + 2
= 100
d, 112 - 102 - 32
= 121 - 100 - 9
= 21 - 9
= 12
e, 13 + 23 + 33 + 43 + 53
= ( 1+ 2+3+4+5)2
= 152
= 225
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
11 tháng 10 2018
a/
\(3S=3+3^2+3^3+3^4+...+3^{120}\)
\(2S=3S-S=3^{120}-1\Rightarrow S=\frac{3^{120}-1}{2}\)
b/ \(S=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(S=13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)
\(S=13+3^3.13+...+3^{117}.13=13\left(1+3^3+...+3^{117}\right)\) chia hết cho 13
c/
\(S=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)
\(S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)
\(S=40+3^4.40+...+3^{116}.40=40\left(1+3^4+...+3^{116}\right)\) chia hết cho 40
29 tháng 8 2023
\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)
\(\dfrac{8}{7}A=\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\)
\(\dfrac{8}{7}A=1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\)
\(\dfrac{8}{7}A=1-\dfrac{1}{89}=\dfrac{88}{89}\Rightarrow A=\dfrac{88}{89}:\dfrac{8}{7}=\dfrac{77}{89}\)
\(B=\dfrac{5^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{37.40}\)
\(B=\dfrac{25}{1.4}+\dfrac{9}{4.7}+\dfrac{9}{7.10}+...+\dfrac{9}{37.40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{37.40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\)
\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{40}=\dfrac{277}{120}\Rightarrow B=\dfrac{277}{120}:\dfrac{1}{3}=\dfrac{277}{40}\)
29 tháng 8 2023
\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)
\(=7\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\right)\)
\(=7\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\right)\)
\(=7.\left(1-\dfrac{1}{89}\right)=7.\dfrac{88}{89}=\dfrac{616}{89}\)
ME
2
1 tháng 10 2018
a)31x32x33x........x3100
=31+2+3+4+...+100
=3(100+1)x(100-1+1):2
=3101x100:2
=35050
Bài b mình không biết làm
10 tháng 8 2019
\(A=3^1+3^2+3^3+3^4+...+3^{199}\)
\(3A=3^2+3^3+3^4+3^5+...+3^{200}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{200}\right)-\left(3^1+3^2+3^3+...+3^{199}\right)\)
\(2A=3^{200}-3^1\)
\(A=\frac{3^{200}-3}{2}\)
=))
XO
10 tháng 8 2019
Đặt \(A=3^1+3^2+3^3+...+3^{199}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{200}\)
Lấy 3A trừ A theo vế ta có :
\(3A-A=\left(3^2+3^3+3^4+..+3^{200}\right)-\left(3^1+3^2+3^3+..+3^{199}\right)\)
\(2A=3^{200}-1\)
\(A=\frac{3^{200}-1}{2}\)
Vậy \(3^1+3^2+3^3+..+3^{199}=\frac{3^{200}-1}{2}\)
LP
0
TA
0
92:33=(32)2:33=34:33=34–3=3
52.252= 52.(52)2=52.54=52+4=56
a)92 : 33 = (32)2 : 33 = 34 : 33 = 3.
b) 52 . 252 = 52 . (52)2 = 52 . 54 = 56.
c) \(\left(\frac{1}{3}\right)^2\) . \(\left(\frac{1}{9.3}\right)^2\) = \(\frac{1^2}{3^2}\). \(\frac{1^2}{27^2}\)= \(\frac{1}{9}\).\(\frac{1}{729}\)= \(\frac{1}{2511}\)