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Số chính phương khi chia 3 chỉ dư 0 hoặc 1.
Trường hợp 1:
\(a^2\equiv1\left(mod3\right);b^2\equiv0\left(mod3\right)\Leftrightarrow a^2+b^2\equiv1\left(mod3\right)\)(loại)
Trường hợp 2:
\(a^2\equiv1\left(mod\right)3;b^2\equiv1\left(mod3\right)\Leftrightarrow a^2+b^2\equiv2\left(mod3\right)\)(loại)
Trường hợp 3:
\(a^2\equiv0\left(mod3\right);b^2\equiv0\left(mod3\right)\Leftrightarrow a^2+b^2\equiv0\left(mod3\right)\) ( thỏa mãn )
Vậy có đpcm.
Giải:
Giả sử a không ⋮ 3 ➩ b không ⋮ 3
➩\(a^2 - 1 + b^2-1\) ⋮ 3
Mà \(a^2 +b^2\)➩2⋮ 3 (không có thể)
Vậy ➩a và b ⋮ 3.
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P = 2.3.4....a => P chia hết cho 3
=> P - 1 : 3 dư 2 => Ko là SCP
Ta có : 3.4.....a lẻ = 2k+1 => P = 2(2k+1) = 4k + 2
=> P + 1 = 4k + 2 + 1 = 4k + 3 : 4 dư 3 => Ko là SCP
=> P - 1 và P + 1 Ko là SCP
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Ta có: \(S=\dfrac{4}{1\cdot3}+\dfrac{16}{3\cdot5}+\dfrac{36}{5\cdot7}+...+\dfrac{2500}{49\cdot51}\)
\(=1+\dfrac{1}{1\cdot3}+1+\dfrac{1}{3\cdot5}+1+\dfrac{1}{5\cdot7}+...+1+\dfrac{1}{49\cdot51}\)
\(=25+\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{49\cdot51}\right)\)
\(=25+\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=25+\dfrac{1}{2}\left(1-\dfrac{1}{51}\right)\)
\(=25+\dfrac{1}{2}\cdot\dfrac{50}{51}\)
\(=25+\dfrac{25}{51}\)
\(=25\cdot\dfrac{52}{51}=\dfrac{1300}{51}\)
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Giả sử tồn tại n thoả mãn đề bài.
Dễ thấy \(2019^{2018}+1\) chẵn nên \(n^3+2018n\), suy ra n chẵn.
Do đó \(n^3+2018n⋮4\).
Mặt khác ta có \(2019^{2018}\equiv\left(-1\right)^{2018}\equiv1\left(mod4\right)\Rightarrow2019^{2018}+1\equiv2\left(mod4\right)\).
Điều này là vô lí vì VT chia hết cho 4 còn VP không chia hết cho 4.
Vậy không tồn tại n thoả mãn đề bài.
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-8/12 rút gọn bằng-2/3; 15/-60 =-1/4; -16/-72=2/9;35/14.15=1/6
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Phần bể chưa có nước bằng:
1 - \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\) (thể tích bể)
Bể sẽ đầy sau:
\(\dfrac{3}{4}\) : \(\dfrac{1}{8}\) = 6 (giờ)
Đs...
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Ta có : p8n+3p4n- 4 = (p4n)2+3p4n- 4
Vì p là số nguyên tố lớn hơn 5 nên p có tận cùng là chữ số 1;3;7 hoặc 9
+) Với p = (...1), ta có: p4n=(...1)4n=(...1)
=> (p4n)2=(...1)2=(...1); 3p4n= 3.(...1)=(...3)
=>(p4n)2+3p4n- 4=(...1)+(...3)-4=(...0) chia hết cho 5
+) Với p = (...3), ta có: p4n=(...3)4n=(...1)
=> (p4n)2=(...1)2=(...1); 3p4n= 3.(...1)=(...3)
=>(p4n)2+3p4n- 4=(...1)+(...3)-4=(...0) chia hết cho 5
+) Với p = (...7), ta có: p4n=(...7)4n=(...1)
=> (p4n)2=(...1)2=(...1); 3p4n= 3.(...1)=(...3)
=>(p4n)2+3p4n- 4=(...1)+(...3)-4=(...0) chia hết cho 5
+) Với p = (...9), ta có: p4n=[(...9)2n]2=(...1)2=(...1)
=> (p4n)2=(...1)2=(...1); 3p4n= 3.(...1)=(...3)
=>(p4n)2+3p4n- 4=(...1)+(...3)-4=(...0) chia hết cho 5
Vậy p8n+3p4n- 4 chia hết cho 5 khi p là số nguyên tố lớn hơn 5
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a + 3 ≤x≤a + 2018 ( a ∈N )
vậy x thuộc (a+3;a+4;a+5;a+6;...;a+2018)
tổng:
a+3+a+4+a+5+a+6+a+7+...+a+2018
=a*2016+3+4+5+6+7+...+2018
=a*2016+(2018+3)*2016:2
-----đến đây cậu làm đc ùi-mik lười lắm ------
\(1)-\left(\dfrac{3}{5}+\dfrac{3}{4}\right)-\left(\dfrac{-3}{4}+\dfrac{2}{5}\right)\\ =-\dfrac{3}{5}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{2}{5}\\ =\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\left(-\dfrac{3}{5}-\dfrac{2}{5}\right)\\ =-1\\ 2)-\left(\dfrac{3}{7}+\dfrac{3}{8}\right)-\left(-\dfrac{3}{8}+\dfrac{4}{7}\right)\\ =-\dfrac{3}{7}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{7}\\ =\left(\dfrac{3}{8}-\dfrac{3}{8}\right)+\left(-\dfrac{3}{7}-\dfrac{4}{7}\right)\\ =-1\\ 3)\left(\dfrac{-25}{27}-\dfrac{31}{42}\right)-\left(\dfrac{-7}{27}-\dfrac{3}{42}\right)\\ =-\dfrac{25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}\\ =\left(\dfrac{-25}{27}+\dfrac{7}{27}\right)+\left(\dfrac{-31}{42}+\dfrac{3}{42}\right)\\ =\dfrac{-18}{27}+\dfrac{-28}{42}\\ =\dfrac{-2}{3}+\dfrac{-2}{3}\\ =-\dfrac{4}{3}\)
Bài 17:
4: \(\left(-\dfrac{1}{2}+\dfrac{7}{6}\right)-\left(-\dfrac{5}{2}+\dfrac{1}{6}\right)\)
\(=-\dfrac{1}{2}+\dfrac{7}{6}+\dfrac{5}{2}-\dfrac{1}{6}\)
=2-1=1
5: \(\left(\dfrac{16}{21}+\dfrac{27}{12}\right)-\left(\dfrac{14}{12}-\dfrac{5}{21}\right)\)
\(=\dfrac{16}{21}+\dfrac{27}{12}-\dfrac{14}{12}+\dfrac{5}{21}\)
\(=1-\dfrac{13}{12}=-\dfrac{1}{12}\)
6: \(\left(\dfrac{13}{23}+\dfrac{-15}{4}\right)+\left(\dfrac{10}{23}-\dfrac{1}{4}\right)\)
\(=\dfrac{13}{23}+\dfrac{10}{23}-\dfrac{15}{4}-\dfrac{1}{4}\)
\(=1-4=-3\)
Bài 18:
1: \(6\dfrac{2}{5}-\left(2\dfrac{4}{9}+4\dfrac{2}{5}\right)\)
\(=6+\dfrac{2}{5}-2-\dfrac{4}{9}-4-\dfrac{2}{5}\)
\(=-\dfrac{4}{9}\)
2: \(7\dfrac{3}{5}-\left(2\dfrac{5}{7}+5\dfrac{3}{5}\right)\)
\(=7+\dfrac{3}{5}-2-\dfrac{5}{7}-5-\dfrac{3}{5}\)
\(=-\dfrac{5}{7}\)
3: \(8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=8+\dfrac{2}{7}-3-\dfrac{4}{9}-4-\dfrac{2}{7}\)
\(=1-\dfrac{4}{9}=\dfrac{5}{9}\)
4: \(8\dfrac{2}{9}-\left(4\dfrac{2}{9}-5\dfrac{1}{2}\right)\)
\(=8+\dfrac{2}{9}-4-\dfrac{2}{9}+5+\dfrac{1}{2}\)
\(=9+\dfrac{1}{2}=\dfrac{19}{2}\)
5: \(21\dfrac{4}{11}-\left(1\dfrac{3}{5}+7\dfrac{4}{11}\right)\)
\(=21+\dfrac{4}{11}-1-\dfrac{3}{5}-7-\dfrac{4}{11}\)
\(=13-\dfrac{3}{5}=\dfrac{62}{5}\)
6: \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}\)
\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)
7: \(\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}\)
\(=5+\dfrac{7}{11}=\dfrac{62}{11}\)
8: \(\left(7\dfrac{8}{9}+2\dfrac{3}{13}\right)-4\dfrac{8}{9}\)
\(=7+\dfrac{8}{9}+2+\dfrac{3}{13}-4-\dfrac{8}{9}\)
\(=5+\dfrac{3}{13}=\dfrac{68}{13}\)
9: \(\left(6\dfrac{5}{7}+2\dfrac{7}{9}\right)-4\dfrac{5}{7}\)
\(=6+\dfrac{5}{7}+2+\dfrac{7}{9}-4-\dfrac{5}{7}\)
\(=4+\dfrac{7}{9}=\dfrac{43}{9}\)