1. tính
a) 1/1.4+1/4.7+1/7.10+...+1/28.31 b) 5/1.3+5/3.5+5/5.7+...+5/99.101
2. so sánh
A=10 mũ 5 +4/10 mũ 5 - 1 và B= 10 mũ 5+3/10 mũ 5 - 2
3. tìm n thuộc Z để giá trị phân số có giá trị nguyên: A= n-2/n+3 ; B=3n+1/n-1
4. tìm x, y thuộc n biết: x/3-4/y=1/5
5. chứng tỏ rằng các phân số sau tối giản với mọi số tự nhiên n
a) n + 1/ 2n + 3 b) 2n +3/ 4n+8
Ai làm đúng mình tick cho!
B1
a)
\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{28\cdot31}\\
=\dfrac{1}{3}\cdot\dfrac{3}{1\cdot4}+\dfrac{1}{3}\cdot\dfrac{3}{4\cdot7}+\dfrac{1}{3}\cdot\dfrac{3}{7\cdot10}+...+\dfrac{1}{3}\cdot\dfrac{3}{28\cdot31}\\
=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{28\cdot31}\right)\\
=\dfrac{1}{3}\cdot\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{28}-\dfrac{1}{31}\right)\\
=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{31}\right)\\
=\dfrac{1}{3}\cdot\dfrac{30}{31}\\
=\dfrac{10}{31}\)
b)
\(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{99\cdot101}\\
=\dfrac{5}{2}\cdot\dfrac{2}{1\cdot3}+\dfrac{5}{2}\cdot\dfrac{2}{3\cdot5}+\dfrac{5}{2}\cdot\dfrac{2}{5\cdot7}+...+\dfrac{5}{2}\cdot\dfrac{2}{99\cdot101}\\
=\dfrac{5}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\\
=\dfrac{5}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\
=\dfrac{5}{2}\cdot\left(1-\dfrac{1}{101}\right)\\
=\dfrac{5}{2}\cdot\dfrac{100}{101}\\
=\dfrac{250}{101}\)
B2
\(A=\dfrac{10^5+4}{10^5-1}=\dfrac{10^5-1+5}{10^5-1}=\dfrac{10^5-1}{10^5-1}+\dfrac{5}{10^5-1}=1+\dfrac{5}{10^5-1}\\
B=\dfrac{10^5+3}{10^5-2}=\dfrac{10^5-2+5}{10^5-2}=\dfrac{10^5-2}{10^5-2}+\dfrac{5}{10^5-2}=1+\dfrac{5}{10^5-2}
\)
Vì \(10^5-1>10^5-2\Rightarrow\dfrac{5}{10^5-1}< \dfrac{5}{10^5-2}\Rightarrow1+\dfrac{5}{10^5-1}< 1+\dfrac{5}{10^5-2}\Leftrightarrow A< B\)
B3
\(A=\dfrac{n-2}{n+3}\)
Để \(A\) có giá trị nguyên thì \(n-2⋮n+3\)
\(n-2=n+3+\left(-5\right)⋮n+3\Rightarrow-5⋮n+3\Rightarrow n+3\inƯ\left(-5\right)\)
\(Ư\left(-5\right)=\left\{-5;-1;1;5\right\}\)
| n+3 | -5 | -1 | 1 | 5 |
| n | -8 | -4 | -2 | 2 |
Vậy \(n\in\left\{-8;-4;-2;2\right\}\)
\(B=\dfrac{3n+1}{n-1}\)Để \(A\) có giá trị nguyên thì \(3n+1⋮n-1\)
\(3n+1=3n-3+4⋮n-1\Leftrightarrow3\cdot\left(n-1\right)+4⋮n-1\Rightarrow4⋮n-1\Rightarrow n-1\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
| n-1 | -4 | -2 | -1 | 1 | 2 | 4 |
| n | -3 | -1 | 0 | 2 | 3 | 5 |
Vậy \(n\in\left\{-3;-1;0;2;3;5\right\}\)
B4
\(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\\ \dfrac{4}{y}=\dfrac{x}{3}-\dfrac{1}{5}\\ \dfrac{4}{y}=\dfrac{5x}{15}-\dfrac{3}{15}\\ \dfrac{4}{y}=\dfrac{5x-3}{15}\\ \Rightarrow4\cdot15=\left(5x-3\right)\cdot y\\ 60=\left(5x-3\right)\cdot y\\ \Rightarrow\left\{{}\begin{matrix}y=60:\left(5x-3\right)\\y\inƯ\left(60\right)\end{matrix}\right.\)
\(Ư\left(60\right)=\left\{1;2;3;4;5;6;10;12;15;20;30;60\right\}\)
| y | 1 | 2 | 3 | 4 | 5 | 6 | 10 | 12 | 15 | 20 | 30 | 60 |
| 5x-3 | 60 | 30 | 20 | 15 | 12 | 10 | 6 | 5 | 4 | 3 | 2 | 1 |
| 5x | 63 | 33 | 23 | 18 | 15 | 13 | 9 | 8 | 7 | 6 | 5 | 4 |
| x | KCĐ | KCĐ | KCĐ | KCĐ | 3 | KCĐ | KCĐ | KCĐ | KCĐ | KCĐ | 1 | KCĐ |
*KCĐ: không chia được (không phải là không chia được, chỉ là x là số tự nhiên nên khi chia kết quả là phân số)
Vậy chỉ có hai cặp \(x;y\) phù hợp là \(\left(3;5\right),\left(1;30\right)\)
B5
a)
\(\dfrac{n+1}{2n+3}\)
Gọi ƯCLN\(\left(n+1,2n+3\right)\) là \(d\left(d\in N^{\circledast}\right)\)
Ta có:
\(n+1⋮d\\ \Rightarrow2\cdot\left(n+1\right)⋮d\\ \Leftrightarrow2n+2⋮d\\ 2n+3⋮d\\ \left(2n+3\right)-\left(2n+2\right)⋮d\\ 1⋮d\\ \Rightarrow d=1\)
Vậy ƯCLN\(\left(n+1,2n+3\right)\) là 1
\(\Rightarrow\dfrac{n+1}{2n+3}\) tối giản
