Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(n\in\left\{-8;-4;-2;2\right\}\)
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 \(n\in\left\{-3;-1;0;2;3;5\right\}\)
bài 1
a, \(A=\frac{3}{x-1}\)
Để A thuộc Z suy ra 3 phải chia hết cho x-1
Suy ra x-1 thuộc ước của 3
Suy ra x-1 thuộc tập hợp -3;-1;1;3
Suy ra x tuộc tập hợp -2;0;2;4
"nếu ko thích thì lập bảng" mấy ccaau kia tương tự
\(a,\)\(1,\)\(A=\frac{3}{x-1}\)
\(A\in Z\Leftrightarrow\frac{3}{x-1}\in Z\)\(\Rightarrow3\)\(⋮\)\(x-1\)
\(\Leftrightarrow x-1\inƯ_3\)
Mà \(Ư_3=\left\{1;3;-1;-3\right\}\)
\(...........\)
\(2,\)\(B=\frac{x-2}{x+3}\)
\(B\in Z\Leftrightarrow\frac{x-2}{x+3}\in Z\)\(\Rightarrow\frac{x+3-5}{x+3}\in Z\)\(\Rightarrow1-\frac{5}{x+3}\in Z\)
\(\Leftrightarrow\frac{5}{x+3}\in Z\)\(\Rightarrow5\)\(⋮\)\(x+3\)
Mà \(Ư_5=\left\{1;5;-1;-5\right\}\)
\(.....\)
\(3,\)\(C=\frac{x^2-1}{x+1}=\frac{\left(x-1\right)\left(x+1\right)}{x+1}=x-1\)
\(C\in Z\Leftrightarrow x-1\in Z\)
\(\Rightarrow x\in Z\)
Gọi d là ƯCLN\((2n-3,3n-5)\)\((d\inℕ^∗)\)
Ta có : \(\hept{\begin{cases}2n-3⋮d\\3n-5⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}3(2n-3)⋮d\\2(3n-5)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}6n-9⋮d\\6n-10⋮d\end{cases}}\)
\(\Rightarrow(6n-10)-(6n-9)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy : ....
a,Tự làm đi bạn nhé
b, \(\frac{x}{4}=\frac{5}{y}\)
\(\Rightarrow x\cdot y=63=1\cdot63=63\cdot1=(-1)(-63)=(-63)(-1)\)
Vậy :....
\(c)\frac{x}{6}=\frac{3}{y}\)
\(\Rightarrow xy=3\cdot6\)
\(\Rightarrow xy=18\)
Tự lập bảng :>