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3 + 3 + 3 - 3 - 3 = 3
3 - 3 - 3 : 3 + 1 = 2
3 + 3 - 3 + 3 : 3 = 2
CÒN LẠI ĐNAG NGHĨ
3 + 3 - 3 - 3 : 3 = 2
3 + 3 - 3 + 3 - 3 = 3
3 + 3 - 3 + 3 : 3 = 4
3 + 3 : 3 + 3 : 3 = 5
x - (2/11.13 + 2/13.15 + 2/15.17 +...+ 2/53.55) = 3/11
=> x - (1/11 - 1/13 + 1/13 - 1/15 + 1/15 - 1/17 +...+ 1/53 - 1/55) = 3/11
=> x - (1/11 - 1/55) = 3/11
=> x - (5/55 - 1/55) = 3/11
=> x - 4/55 = 15/55
=> x = 15/55 + 4/55
=> x = 19/55
1) So sánh
3 77/379 và 3 79/381
2)
A= 1/6 + 1/10 + 1/15 + 1/21 + 1/28 + 1/36
Giúp mình nhé❤❤❤❤❄▫〰▫▫▫▫▫▫
2) A = \(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}\)
=> \(\frac{1}{2}\).A = \(\frac{1}{2}\).\(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}\right)\)
=> \(\frac{1}{2}\).A = \(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
=> \(\frac{1}{2}\).A = \(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
=> \(\frac{1}{2}\).A = \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
=> \(\frac{1}{2}\).A = \(\frac{1}{3}-\frac{1}{9}\)
=> \(\frac{1}{2}\).A = \(\frac{2}{9}\)
=> A = \(\frac{2}{9}:\frac{1}{2}\)
=> A = \(\frac{4}{9}\)
a) A = 4 + 4² + 4³ + ... + 4¹²
= 4.(1 + 4 + 4² + 4³ + ... + 4¹¹) ⋮ 4
Vậy A ⋮ 4
b) A = 4 + 4² + 4³ + 4⁴ + ... + 4¹²
= (4 + 4²) + (4³ + 4⁴) + ... + (4¹¹ + 4¹²)
= 4.(1 + 4) + 4³.(1 + 4) + ... + 4¹¹.(1 + 4)
= 4.5 + 4³.5 + ... + 4¹¹.5
= 5.(4 + 4³ + ... + 4¹¹) ⋮ 5
Vậy A ⋮ 5
c) A = 4 + 4² + 4³ + 4⁴ + ... + 4¹²
= (4 + 4² + 4³) + (4⁴ + 4⁵ + 4⁶) + ... + (4¹⁰ + 4¹¹ + 4¹²)
= 4.(1 + 4 + 4²) + 4⁴.(1 + 4 + 4²) + ... + 4¹⁰.(1 + 4 + 4²)
= 4.21 + 4⁴.21 + ... + 4¹⁰.21
= 21.(4 + 4⁴ + ... + 4¹⁰) ⋮ 21
Vậy A ⋮ 21
Bạn tham khảo bài này xong tự làm nha :
So sánh 2301và 3201
Ta có : 2301=2200.2=(23)100.2=8100.2
:3201=3200.3= ( 32)100=9100.3
Do 8<9=>8100<9100 :2<3 => 8100.2<9100.3=>2301<3201
\(A=2+2^2+2^3+.....+2^{2023}+2^{2024}+2^{2025}\)
\(A=2\left(1+2+4\right)+.....+2^{2023}\left(2+2+4\right)\)
\(A=2.7+.....+2^{2023}.7⋮7\left(đpcm\right)\)
\(A=2+2^2+2^3+...+2^{2025}\\ A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2023}+2^{2024}+2^{2025}\right)\\ A=2.\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2023}\left(1+2+2^2\right)\\ A=2.7+2^4.7+2^7.7+...+2^{2023}.7\\ A=\left(2+2^4+2^7+...+2^{2023}\right).7⋮7\)
Vậy \(A⋮7\left(đpcm\right)\)