\(B=\frac{1}{3\cdot5}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{95\cdot96}\)

\(=\frac{1}{15}+\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{95\cdot96}\right)\)

\(=\frac{1}{15}+\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{95}-\frac{1}{96}\right)\)

\(=\frac{1}{15}+\left(\frac{1}{4}-\frac{1}{96}\right)\)

\(=\frac{1}{15}+\frac{23}{96}=\frac{49}{160}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{37\cdot39}\)

\(\frac{1}{2}A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{37\cdot39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)

\(=\frac{1}{2}\cdot\frac{4}{13}\)

\(\Rightarrow A=\frac{4}{13}\)

• nhattquynhh cảm ơn bạn nhiều nhé

Đặt \(d=\left(x+1,2021x+2020\right)\).

Suy ra 

\(\hept{\begin{cases}x+1⋮d\\2021x+2020⋮d\end{cases}}\Rightarrow\hept{\begin{cases}2021x+2021⋮d\\2021x+2020⋮d\end{cases}}\Rightarrow\left(2021x+2021\right)-\left(2021x+2020\right)=1⋮d\)

suy ra \(d=1\).

Suy ra đpcm. 

A=4.(1/1.2+1/2.3+1/3.4+...+1/2014.2015)

A=4.(1-1/2+1/2-1/3+1/3-1/4....1/2014-1/2015)

A=4.(1-1/2015)

A=4.2014/2015

A=8056/2015

Xong rồi đó bạn nhé nhé

a) \(B=\frac{x-2}{x+3}=\frac{x+3-5}{x+3}=1-\frac{5}{x+3}\)là số nguyên tương đương với \(\frac{5}{x+3}\)là số nguyên

tương đương \(x+3\inƯ\left(5\right)=\left\{-5,-1,1,5\right\}\Leftrightarrow x\in\left\{-8,-4,-2,2\right\}\). 

b) \(C=\frac{2x+1}{x-3}=\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\inℤ\Leftrightarrow\frac{7}{x-3}\inℤ\)

\(\Leftrightarrow x-3\inƯ\left(7\right)=\left\{-7,-1,1,7\right\}\Leftrightarrow x\in\left\{-4,2,4,10\right\}\).

c) \(D=\frac{x^2-1}{x+1}=\frac{x^2+x-x-1}{x+1}=x-1\)

Vậy \(D\inℤ\Leftrightarrow x\inℤ\backslash\left\{-1\right\}\).

TIỂU HỌC MÀ THẢ THÍNH HAY NHỈ ?

CẦN VÉ BÁO CÁO HẢ BN , MINK CHO FREE LUÔN NÀY

a) Vì \(\widehat{xOy}\)và \(\widehat{yOz}\)là hai góc kề bù:

\(\Rightarrow\widehat{yOz}+\widehat{xOy}=180^o\)

Thay \(\widehat{xOy}=130^o\), ta có:

\(130^o+\widehat{yOz}=180^o\)

\(\Rightarrow\widehat{yOz}=180^o-130^o\)

\(\Rightarrow\widehat{yOz}=50^o\)

b) Trên nửa mặt phẳng bờ chứa tia Ox, ta có:

\(\widehat{xOt}< \widehat{xOy}\left(80^o< 130^o\right)\)

\(\Rightarrow Ot\)nằm giữa \(Ox\)và \(Oy\)

\(\Rightarrow\widehat{yOt}+\widehat{xOt}=\widehat{xOy}\)

Thay \(\widehat{xOt}=80^o;\widehat{xOy}=130^o\)ta có:

\(80^o+\widehat{yOt}=130^o\)

\(\Rightarrow\widehat{yOt}=130^o-80^o\)

\(\Rightarrow\widehat{yOt}=50^o\)

\(\Rightarrow\widehat{tOy}=\widehat{yOz}=50^o\)(1)

c) Vì \(\widehat{tOy}=\widehat{yOz}=50^o=\frac{1}{2}\widehat{tOz=100^o}\)

\(\Rightarrow Oy\)nằm giữa \(Ox\)và \(Ot\)(2)

Từ (1) và (2) \(\Rightarrow Oy\)là tia phân giác của \(\widehat{tOz}\)