3×2=?
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a: A=3^2(1^2+2^2+...+10^2)
=9*385
=3465
b: B=2^3(1^3+2^3+...+10^3)
=8*3025
=24200
\(2A-A=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+...+2^{20}\right)\)
\(A=2^{21}-2\)
B tương tự câu A
\(5C-C=\left(5^2+5^3+...+5^{51}\right)-\left(5+5^2+...+5^{50}\right)\)
\(C=\dfrac{5^{51}-5}{4}\)
\(3D-D=3+3^2+...+3^{101}-\left(1+3+...+3^{100}\right)\)
\(D=\dfrac{3^{101}-1}{2}\)
\(A=2^1+2^2+2^3+...+2^{20}\)
\(2\cdot A=2^2+2^3+2^4+...+2^{21}\)
\(A=2^{21}-2\)
\(B=2^1+2^3+2^5+...+2^{99}\)
\(4\cdot B=2^3+2^5+2^7+...+2^{101}\)
\(B=\)\(\left(2^{101}-2\right):3\)
\(C=5^1+5^2+5^3+...+5^{50}\)
\(5\cdot C=5^2+5^3+5^4+...+5^{51}\)
\(C=(5^{51}-5):4\)
\(D=3^0+3^1+3^2+...+3^{100}\)
\(3\cdot D=3^1+3^2+3^3+...+3^{101}\)
\(D=(3^{101}-1):2\)
a) \(2^5\cdot2^7\)
\(=2^{5+7}\)
\(=2^{12}\)
b) \(2^3\cdot2^2\)
\(=2^{3+2}\)
\(=2^5\)
c) \(2^4\cdot2^3\cdot2^5\)
\(=2^{4+3+5}\)
\(=2^{12}\)
d) \(2^2\cdot2^4\cdot2^6\cdot2\)
\(=2^{2+4+6+1}\)
\(=2^{13}\)
e) \(2\cdot2^3\cdot2^7\cdot2^4\)
\(=2^{1+3+7+4}\)
\(=2^{15}\)
f) \(3^8\cdot3^7\)
\(=3^{8+7}\)
\(=3^{15}\)
g) \(3^2\cdot3\)
\(=3^{2+1}\)
\(=3^3\)
h) \(3^4\cdot3^2\cdot3\)
\(=3^{4+2+1}\)
\(=3^7\)
I) \(3\cdot3^5\cdot3^4\cdot3^2\)
\(=3^{1+5+4+2}\)
\(=3^{12}\)
Lời giải chi tiết
12 1 13 12 – 02 (0 + 1)2 02 +12
22 1 + 3 23 32 – 12 (1 + 2)2 12 + 22
32 1 + 3 + 5 33 62 – 32 (2 + 3)2 22 + 32
43 102 – 62
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\(\frac{2}{3}\times5+\frac{2}{3}\times7+\frac{2}{3}\times9+\frac{2}{3}\times11+\frac{2}{3}\times13+\frac{2}{3}\times15\)
=\(\frac{2}{3}\)\(\times\)( 5+7+9+11+15)
= \(\frac{2}{3}\)\(\times\)47
= \(\frac{94}{3}\)
Có friend forever II Lê Tiến Đạt giải rồi nhé nên đừng bắt tui giải nữa ( chuồn là thượng sách)
a. M=-1^2+2^2-3^2+4^2-...-99^2+100^2.
M=(2-1)(2+1)+(4-3)(4+3)+...+(100-99)(100+99)
M=3+7+...+199
=>2M=3+7+...+199+3+7+...+199 (198 số)
=(3+199)+(7+195)+...+(199+3) (99 cặp)
=202.99
=19998
=>M=19998:2=9999
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A= 1 - 2 - 22 - 23 + 24 +...+ 22022 (sửa đề)
= -13 + (24 + 25 + 26 + ... + 22022)
2A = -26 + (25 + 26 + 27 + ... + 22023)
2A - A = -26 + (25 + 26 + 27 + ... + 22023) - [-13 + (24 + 25 + 26 + ... + 22022)]
A = -13 +(22023 - 24)
= 22023 - 29
Vậy...
B = 1 + 3 + 32 + 33 + 34 + ... + 32022
3B = 3 + 32 + 33 + 34 + 35 +...+ 32023
3B - B = 3 + 32 + 33 + 34 + 35 +...+ 32023 - (1 + 3 + 32 + 33 + 34 + ... + 32022)
2B = 32023 - 1
=> B = \(\dfrac{3^{2023}-1}{2}\)
Vậy...
#Ayumu
3 x 2 = 6
3x2=6