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a) 15 + 23 = 1 + 8 = 9 = 32 ( là số chính phương )
b) 52 + 122 = 25 + 144 = 169 = 132 ( là số chính phương )
c) 26 + 62 = 64 + 36 = 100 = 1002 ( là số chính phương )
d) 13 + 23 + 33 + 43 + 53 + 63
= 1 + 8 + 27 + 64 + 125 + 216
= 441 = 212 ( là số chính phương )
a) 15 + 23=1 + 8 = 9 (là số chính phương)
b) 52 + 122= 25 + 144= 169 (là số chính phương)
c) 26 + 62= 64 + 36=100 (là số chính phương)
d) 142 – 122= 196 - 144=52 (không là số chính phương)
e) 13 + 23 + 33 + 43 + 53 + 63= 1 + 8 + 27 + 64 + 125 + 216 = 411 (là số chính phương)
i) \(2345-1000\div\left[19-2\left(21-18\right)^2\right]\)
\(=\)\(2345-1000\div\left[19-2.3^2\right]\)
\(=\)\(2345-1000\div\left[19-2.9\right]\)
\(=\)\(2345-1000\div\left[19-18\right]\)
\(=\)\(2345-1000\div1\)
\(=\)\(2345-1000\)
\(=\)\(1345\)
j) \(128-\left[68+8\left(37-35\right)^2\right]\div4\)
\(=\)\(128-\left[68+8.2^2\right]\div4\)
\(=\)\(128-\left[68+8.4\right]\div4\)
\(=\)\(128-\left[68+32\right]\div4\)
\(=\)\(128-100\div4\)
\(=\)\(128-25\)
\(=\)\(3\)
k) \(568-\left\{5\left[143-\left(4-1\right)^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-3^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-9\right]+10\right\}\div10\)
\(=\)\(568-\left\{5.134+10\right\}\div10\)
\(=\)\(568-\left\{670+10\right\}\div10\)
\(=\)\(568-680\div10\)
\(=\)\(568-68\)
\(=\)\(500\)
a) \(107-\left\{38+\left[7.3^2-24\div6+\left(9-7\right)^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.3^2-24\div6+2^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.9-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[63-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+67\right\}\div15\)
\(=\)\(107-105\div15\)
\(=\)\(107-7\)
\(=\)\(7\)
b) \(307-\left[\left(180-160\right)\div2^2+9\right]\div2\)
\(=\)\(307-\left[20\div4+9\right]\div2\)
\(=\)\(307-\left[5+9\right]\div2\)
\(=\)\(307-14\div2\)
\(=\)\(307-7\)
\(=\)\(300\)
c) \(205-\left[1200-\left(4^2-2.3\right)^3\right]\div40\)
\(=\)\(205-\left[1200-\left(16-6\right)^3\right]\div40\)
\(=\)\(205-\left[1200-10^3\right]\div40\)
\(=\)\(205-\left[1200-1000\right]\div40\)
\(=\)\(205-200\div40\)
\(=\)\(205-5\)
\(=\)\(200\)
Answer:
a)
Có:
\(15=1.15=\left(-1\right).\left(-15\right)=3.5=\left(-3\right).\left(-5\right)\)
| x-3 | 1 | 15 | -1 | -15 | 3 | 5 | -3 | -5 |
| y+1 | 15 | 1 | -15 | -1 | 5 | 3 | -5 | -3 |
| x | 4 | 18 | 2 | -12 | 6 | 8 | 0 | -2 |
| y | 14 | 0 | -16 | -2 | 4 | 2 | -6 | -4 |
b)
Có:
\(M=1+3+3^2+3^3+3^4+...+3^{98}+3^{99}+3^{100}\)
\(=\left(1+3\right)+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(=4+3^2.\left(1+3+3^2\right)+...+3^{98}.\left(1+3+3^2\right)\)
\(=4+3^2.13+3^{98}.13\)
\(=4+13.\left(3^2+...+3^{98}\right)\)
Vậy M chia 13 dư 4
Có:
\(M=1+3+3^2+3^3+3^4+...+3^{99}+3^{100}\)
\(=1+\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=1+3.40+3^5.40+...+3^{97}.40\)
\(=1+40.\left(3+3^5+...+3^{97}\right)⋮40\)
Mà \(40\left(3+3^5+...+3^{97}\right)⋮40\)
Vậy M chia 40 dư 1
Bài 2*: So sánh:
a) A = 20 + 21 + 22 + 23 + … + 22010
2A = 2(1 + 21+ 22+..+22010)
2A = 21+22+23+..+22011
A = 22011- 20
A = 22011-1
=> A = B
b) A = 2009 . 2011
A = 2009 (2010 + 1 )
A = 2009 . 2010+ 2009 . 1
A = 2010 . 2009 + 2009
B = 20102
B = 2010. 2010
B = 2010 . (2009 + 1)
B = 2010 . 2009 + 2010 . 1
B = 2010 . 2009 + 2010
Ta thấy 2010 > 2009 nên 2010 . 2009 + 2010 > 2010 . 2009 + 2009
c) A = 1030 = (103)10 = 100010
B = 2100 = (210)10 = 102410
=> A < B
d) A = 333444= (3334)111 = (1114.81)111
B = 444333= (4443)111 = (1113.64)111
=> A > B
e) A = 3450 = 33.150 = (33)150= 27150
B = 5300 = 53.100= (53)150 = 15150
=> A > B
*à xin lỗi cho mình sửa chỗ B câu cuối*
B 5450 = (52)150 = 25150
=> A > B
a) \(B=2012.2014=\left(2013-1\right)\left(2013+1\right)=2013.2013-2013+2013-1\)
\(=2013.2013-1< 2013.2013=A\)
b) \(A=2+2^2+2^3+...+2^{2014}\)
\(2A=2^2+2^3+2^4+...+2^{2015}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2015}\right)-\left(2+2^2+2^3+...+2^{2014}\right)\)
\(A=2^{2015}-2< 2^{2015}=B\)
a. S = 1 + 2 + 2^2 + 2^3 + ... + 2^8 + 2^9
Ta có: 2 = 1 . 2
2^2 = 2 . 2
2^3 = 2^2 . 2
.....
=> 1 + 2 + 2^2 + ... + 2^8 + (2^8 . 2)
=> 1 + 2 + 2^2 + ... + (2^8 . 3)
=> 1 + 2 + 2^2 + ... + 2^7 + (2^7 .6)
=> 1 + 2 + 2^2 + ... + (2^7 . 7)
=> .....
=> 1 + 2 . 311
a) A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰²²
2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²³
A = 2A - A
= (2 + 2² + 2³ + 2⁴ + ... + 2²⁰²³) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰²²)
= 2²⁰²³ - 2⁰
= 2²⁰²³ - 1
Vậy A = B
b) A = 2021 . 2023
= (2022 - 1).(2022 + 1)
= 2022.(2022 + 1) - 2022 - 1
= 2022² + 2022 - 2022 - 1
= 2022² - 1 < 2022²
Vậy A < B