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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) \(205-\left[1200-\left(4^2-2.3\right)^3\right]\div40\)
\(=\)\(205-\left[1200-\left(16-6\right)^3\right]\div40\)
\(=\)\(205-\left[1200-10^3\right]\div40\)
\(=\)\(205-\left[1200-1000\right]\div40\)
\(=\)\(205-200\div40\)
\(=\)\(205-5\)
\(=\)\(200\)
\(A=3+3^2+3^3+...+3^{2020}\)
\(3A=3^2+3^3+3^4+...+3^{2021}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{2021}\right)-\left(3+3^2+3^3+...+3^{2020}\right)\)
\(2A=3^{2021}-3\)
\(A=\frac{3^{2021}-3}{2}\)
Từ đây cũng suy ra \(x=2021\).
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
a) 3.52 + 15.22 - 26 : 2
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) 2021 + 5.[300 - (17 - 7)2]
= 2021 + 5.[300 - 102]
= 2021 + 5.[300 - 100]
= 2021 + 5.200
= 2021 + 1000
= 3021
c) 32.5 + 23.10 - 81
= 9.5 + 8.10 - 81
= 45 + 80 - 81
= 125 - 81
= 44
a) 3.52 + 15.22 - 26 : 2
= 3. 25 + 15 . 4 - 26 : 2
= 75 + 60 - 13
= 135 - 13
= 122
b) 2021 + 5[300 - (17 - 7)2 ]
= 2021 + 5[300 - 102 ]
=2021 + 5[300 - 100]
=2021 + 5.200
=2021 + 1000
= 3021
c) 32 . 5 + 23 . 10 - 81
= 9 . 5 + 8 . 10 - 81
= 45 + 80 - 81
= 125 - 81
= 44