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1-3-5+7+9-11-13+15+...+2017-2019-2021+2023=
=(1-3-5+7)+(9-11-13+15)+...+(2017-2019-2021+2023)=
=0+0+.....+0=0
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\(-\frac{1}{13}+\frac{15}{21}+\frac{\left(-12\right)}{13}+\frac{2}{7}+\frac{2020}{2021}\)
\(=-\frac{1}{13}+\frac{15}{21}-\frac{12}{13}+\frac{6}{21}+\frac{2020}{2021}\)
\(=\left(-\frac{1}{13}-\frac{12}{13}\right)+\left(\frac{15}{21}+\frac{6}{21}\right)+\frac{2020}{2021}\)
\(=-1+1+\frac{2020}{2021}=\frac{2020}{2021}\)
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1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
2x+1 | 1 | 3 | 7 | 21 |
x | 0 | 1 | 3 | 10 |
TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.
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Lời giải:
$H=(1+7)+(7^2+7^3)+(7^4+7^5)+....+(7^{2020}+7^{2021})$
$=(1+7)+7^2(1+7)+7^4(1+7)+....+7^{2020}(1+7)$
$=(1+7)(1+7^2+7^4+....+7^{2020})$
$=8(1+7^2+7^4+....+7^{2020})\vdots 8$
Ta có đpcm.
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Sửa đề:
B = 1 - 3 - 5 + 7 + 9 - 11 - 13 + 15 + ... + 2019 - 2021 - 2023 + 2025 + 2027
= (1 - 3 - 5 + 7) + (9 - 11 - 13 + 15) + ... + (2019 - 2021 - 2023 + 2025) + 2027
= 0 + 0 + ... + 0 + 2027
= 2027
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Trả lời :
\(\left(\frac{1}{4}+0,75\right)\div\frac{5}{7}+\frac{2}{5}+\left|2021\right|\)
\(=\left(\frac{1}{4}+\frac{3}{4}\right)\div\frac{5}{7}+\frac{2}{5}+2021\)
\(=1\div\frac{5}{7}+\frac{2}{5}+2021\)
\(=\frac{7}{5}+\frac{2}{5}+2021\)
\(=\frac{9}{5}+2021\)
\(=\frac{10114}{5}\)
Số hơi xấu, check lại đề.
TL ;
72 - 52 + 2 X ( 13 + 20210 )
= 72 - 52 + 21 x ( 131 + 2021 )
= 2 + 21 x 2033
= 41 x 2033
= 4 x 2033
= 8132
72 - 52 + 2 . ( 13 + 20210 )
= 49 - 25 + 2 . ( 13 + 1 )
= 49 - 25 + 2 . 14
= 49 - 25 + 28
= 52