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giả sử tồn tại các số nguyên t/m:
abc+a=1333.............
xét từng điều kiện ta có
abc+a=a(bc+1)=1333
abc+b=b(ac+1)=1335
abc+c=c(ab+1)=1341
chỉ có 2 số lẻ mới là tích của 1 số lẻ=>a,b,c lẻ=>abc lẻ
=>abc+a chẵn khác 1333(số lẻ)
CM tương tụ vs 2 th khác
=> ko tồn tại a,b,c thỏa mãn
\(A=\)\(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)
\(A=\frac{1}{7}-\frac{1}{103}\)
\(A=\frac{96}{721}\)
\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(B=2\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)
\(3B=2.3\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)
\(3B=2\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)
\(3B=2\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(3B=2\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(3B=2.\frac{96}{721}\)
\(3B=\frac{192}{721}\)
\(\Rightarrow B=\frac{192}{721}:3\)
\(B=\frac{64}{721}\)
\(A=\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)
\(A=\frac{1}{7}-\frac{1}{103}\)
\(A=\frac{96}{721}\)
Vậy \(A=\frac{96}{721}\)
\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(B=\frac{2}{3}.\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)
\(B=\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}.\frac{96}{721}\)
\(B=\frac{64}{721}\)
Vậy \(B=\frac{64}{721}\)
_Chúc bạn học tốt_
a, -(-5)-(+7)+(+3)+(-8)
= 5 - 7 + 3 + (-8)
= (-2) + 3 + (-8)
= 1 + (-8)
= -7
b, -(-15)-|-10|+|-9|-|5I
= 15 - 10 + 9 - 5
= 5 + 9 - 5
= 14 - 5
= 9
c, 14-(-13)-(17)+(-12)
= 14 + 13 - 17 + (-12)
= 27 - 17 + (-12)
= 10 + (-12)
= 2
d, -|-14| + |-10| - (-12) + (-8)
= -14 + 10 + 12 + (-8)
= (-4) + 12 + (-8)
= 7 + (-8)
= -1
e, -(-11) + (-15) + (13) - 21
-x + 20 = - (-15) - (8) + 13
-x + 20 = 15 - 8 + 13
-x + 20 = 7 + 13
- x + 20 = 20
x = 20 - 20
x = 0
-(-10) + x = -13 + (-9) + (-6)
10 + x = -13 - 9 - 6
10 + x = -28
x = -28 - 10
x = -38
\(\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{100.103}\)
\(=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{100}-\frac{1}{103}\)
\(=\frac{1}{7}-\frac{1}{103}\)
\(=\frac{96}{721}\)
\(\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(=\frac{2}{3}.\frac{96}{721}\)
\(=\frac{64}{721}\)
a) Ta có : 51n=\(\overline{...1}\)
47102=472.(474)25=\(\left(\overline{...9}\right).\left(\overline{...1}\right)=\overline{...9}\)
\(\Rightarrow51^n+47^{102}=\left(\overline{...1}\right)+\left(\overline{...9}\right)=\overline{...0}⋮10\)
Vậy 51n+47102\(⋮\)10.
b) Ta có : \(17^5=17.17^4=17.\left(\overline{...1}\right)=\overline{...7}\)
\(24^4=\overline{...6}\)
\(13^{21}=13.\left(13^4\right)^5=13.\left(\overline{...1}\right)=\overline{...3}\)
\(\Rightarrow17^5+24^4-13^{21}=\left(\overline{...7}\right)+\left(\overline{...6}\right)-\left(\overline{...3}\right)=\overline{...0}⋮10\)
Vậy 175+244+1321\(⋮\)10
\(=13\cdot70\cdot\left(\frac{1}{7}+\frac{1}{10}\right)=13\cdot\left(\frac{70}{7}+\frac{70}{10}\right)=13\cdot\left(10+7\right)=13\cdot17=221.\)
Vậy....
=70x13/7+70x13/10
=10x13+7x13
=130+91
=221