Bài 1:
a) Phân tích các số sau ra thừa số nguyên tố: A = 2024 và B = 8 mũ 5 x 125 mũ 6
b) Tìm x, biết: x e Ư (84), x e B (21)
c) Tìm chữ số x, biết: (2x785 + 1500 mũ 11) chia hết cho 15.
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\(\dfrac{\dfrac{\dfrac{9}{9}}{\dfrac{9}{9}}}{\dfrac{\dfrac{9}{9}}{\dfrac{9}{9}}}\) + \(\dfrac{\dfrac{\dfrac{9}{9}}{\dfrac{9}{9}}}{\dfrac{\dfrac{9}{9}}{\dfrac{9}{9}}}\) = 1 + 1 = 2
\(2x-124=x+224\\ \Rightarrow2x+x=224+124\\ \Rightarrow3x=348\\ \Rightarrow x=348:3\\ \Rightarrow x=116\)
2\(x\) - 124 = \(x\) + 224
2\(x\) - \(x\) = 224 + 124
\(x\) = 348
\(\left(1^2+2^2+3^2+...+1000000^2\right).\left(91-273:3\right)\\ =\left(1^2+2^2+3^2+...+1000000^2\right).0=0\)
(12+22+32+..+10000002).(91-273:3)
=(12+22+32+..+10000002).(91-91)
=(12+22+32+..+10000002).0
=0
\(\overline{ab}\) \(\times\) 101 = \(\overline{ab}\) \(\times\)(100 + 1) = \(\overline{ab00}\) + \(\overline{ab}\) = \(\overline{abab}\)
\(A=1+2+2^2+...+2^{2018}\)
\(2A=2+2^3+2^4+...+2^{2019}\)
\(A=2A-A=1-2^{2019}\)
\(B-A=2^{2019}-\left(1-2^{2019}\right)\)
\(B-A=2^{2019}-1+2^{2019}\)
\(B-A=1\)
`#3107`
\(A=1+2+2^2+2^3+...+2^{2018}\) và \(B=2^{2019}\)
Ta có:
\(A=1+2+2^2+2^3+...+2^{2018}\)
\(2A=2+2^2+2^3+...+2^{2019}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2019}\right)-\left(1+2+2^2+2^3+...+2^{2018}\right)\)
\(A=2+2^2+2^3+...+2^{2019}-1-2-2^2-2^3-...-2^{2018}\)
\(A=2^{2019}-1\)
Vậy, \(A=2^{2019}-1\)
Ta có:
\(B-A=2^{2019}-2^{2019}+1=1\)
Vậy, `B - A = 1.`
\(25\cdot32=25\cdot4\cdot8\)
\(=\left(25\cdot4\right)\cdot8=100\cdot8\)
\(=800\)
\(49\cdot101=49\cdot\left(100+1\right)\)
\(=49\cdot100+49\)
\(=4900+49\)
\(=4949\)
(36 x 15 x 169):(5 x 18 x 13)
= (18 x 2 x 3 x 5 x 13 x 13):(5 x 18 x 13)
= 2x3x13x(5 x 18 x 13):(5 x 18 x 13)
= 2 x 3 x 13
= 78
a) 2016 - 100.(x+31)=27:23
<=> 2016 - 100.(x+31) = 24 = 16
<=> 100.(x+31) = 2016 - 16 = 2000
<=> x+31 = 2000:100 = 20
<=> x = 20 - 31 = -11
b) (x+8)3 = 125
<=> (x+8)3 = 53
<=> x+8 = 5
<=> x = 5 - 8
<=> x = -3
\(a,A=2024=2^3\times11\times23\\B=8^5\times 125^6=\left(2^3\right)^5\times\left(5^3\right)^6=2^{15}\times5^{18}\\ b,Ư\left(84\right)=\left\{1;2;3;4;6;7;12;14;21;28;42;84\right\}\\\Rightarrow x\in\left\{1;2;3;4;6;7;12;14;21;28;42;84\right\}\\ x\in B\left(21\right)=\left\{0;21;42;63;84;105;126;147;168;189;210;....\right\}\)