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A=1.1+2.2+3.3+.....+100.100
A=1.(2-1)+2.(3-1)+.......+100.(101-1)
A=1.2+2.3+......+100.101-1-2-3-4-.......-100
3A=1.2.(3-0)+2.3.(4-1)+......+100.101.(102-99)-(1+2+3+....+100).3
3A=1.2.3+2.3.4+....+100.101.102-1.2.3-2.3.4-.....-99.100.101-(1+2+3+......+100).3
3A=100.101.102-101.100.3
3A=101.100.(102-3)
3A=101.100.99
A=101.100.33
A=(mấy tự tính)
\(6⋮\left(x-1\right)\)
\(\Rightarrow x-1\in\text{ư}\left(6\right)=1.2.3.6\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x-1=2\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x-1=6\)
\(x=1+1=2\)
\(x=1+2=3\)
\(x=1+3=4\)
\(x=1+6=7\)
=> x - 1 \(\in U\left[6\right]\in\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Rightarrow x\in\left\{-5;-2;-1;0;2;3;4;7\right\}\)
\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(=\frac{\frac{101.102}{2}}{51}\)
\(=101\)
Bài 1
a/ \(ab+ba=10a+b+10b+a=11a+11b=11\left(a+b\right)\) chia hết cho 11
b/ \(ab-ba=10a+b-10b-a=9a-9b=9\left(a-b\right)\) chia hết cho 9
Bài 2
a/ \(\overline{abcd}=100.\overline{ab}+\overline{cd}=100.\overline{ab}+100.\overline{cd}-99.\overline{cd}=100\left(\overline{ab}+\overline{cd}\right)-99.\overline{cd}\)
Ta có \(\overline{ab}+\overline{cd}\) chia hết cho 99 \(\Rightarrow100\left(\overline{ab}+\overline{cd}\right)\) chia hết cho 99 và \(99.\overline{cd}\) chia hết cho 99 \(\Rightarrow100\left(\overline{ab}+\overline{cd}\right)-99.\overline{cd}\) chia hết cho 99 nên \(\overline{abcd}\) chia hết cho 99
b/ \(\overline{abcdef}=1000.\overline{abc}+\overline{def}=999.\overline{abc}+\left(\overline{abc}+\overline{def}\right)=27.37.\overline{abc}+\left(\overline{abc}+\overline{def}\right)\)
\(\Rightarrow\overline{abcdef}\) chia heets cho 37
Bài 3
a/ \(A=\left(1+3+3^2\right)+...+3^{1998}\left(1+3+3^2\right)=13.\left(1+...+3^{1998}\right)\) chia hết cho 13
b/ \(B=\left(1+4+4^2\right)+...+4^{2010}\left(1+4+4^2\right)=21.\left(1+...+4^{2010}\right)\) chia hết cho 21
A = -1 + -2 + -3 + -4 + ... + -99 + -100
= - ( 1 + 2 +3 + ... + 100)
= - 5050
\(...\\ A=-\left(1+2+3+...+100\right)\\ A=-\left(\frac{\left(1+100\right).100}{2}\right)\\ A=-101.50=-5050\)
Chúc bạn học tốt!!!