\(6+6^2+\cdot\cdot\cdot+6^{10}\)

\(=6\cdot\left(1+6\right)+6^3\cdot\left(1+6\right)+\cdot\cdot\cdot+6^9\cdot\left(1+6\right)\)

\(=6\cdot7+6^3\cdot7+\cdot\cdot\cdot+6^9\cdot7\)

\(=7\cdot\left(6+6^3+\cdot\cdot\cdot+6^9\right)⋮7\)

\(\Rightarrow6+6^2+\cdot\cdot\cdot\cdot+6^{10}⋮7\)

\(5^1-5^9+5^8=5\left(1-5^8+5^7\right)⋮7\Leftrightarrow5^8-5^7-1⋮7\)

\(5\equiv-2\left(mod7\right)\Rightarrow5^3\equiv-1\left(mod7\right)\Rightarrow5^8\equiv4\left(mod7\right);5^7\equiv-2\left(mod7\right)\)

\(5^8-5^7-1\equiv5\left(mod7\right):v\)

\(6+6^2+\cdot\cdot\cdot+6^{10}\)

\(=6\cdot7+6^3\cdot7+\cdot\cdot\cdot+6^9\cdot7\)

\(=7\cdot\left(6+\cdot\cdot\cdot+6^9\right)\)

\(⋮7\)

a) A=1-2-3+4+5-6-7+.....+1996+1997-1998-1999+2000

=(1-2-3+4)+(5-6-7+8)+...+(1997-1998-1999+2000)

=0

b) B=1-3+5-7+....+2001-2003+2005

=(1-3)+(5-7)+...+(2001-2003)+2005

=-2.501+2005

=-1002+2005

=1003

c) C=1-2-3+4+5-6-7+8+.....+1993-1994-1995+1996+1997

=(1-2-3+4)+(5-6-7+8)+...+(1993-1994-1995+1996)+1997

=1997

d) D=1000+998+996+......+10-999-997-995-...-11

=(1000-999)+(998-997)+(996-995)+....+(12-11)+10

=1.495+10

=595

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-1-\frac{1}{2}-...-\frac{1}{5}\)

\(=\frac{1}{6}+\frac{1}{7}+...+\frac{1}{10}\left(đpcm\right)\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\)\(\frac{1}{10}\)

\(A=\frac{1}{1}+\frac{1}{3}+...+\frac{1}{9}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{10}\)

\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}-2.\frac{1}{2}-2.\frac{1}{4}-...-2.\frac{1}{10}\)

\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}-1-\frac{1}{2}-...-\frac{1}{5}\)

\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\left(đpcm\right)\)

                                    ~~~Hok tốt~~~

\(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{10}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{10}-\left(1+\frac{1}{2}+...+\frac{1}{5}\right)\)

\(=\frac{1}{6}+\frac{1}{7}+...+\frac{1}{10}\)

Ta có :

\(\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\)

\(=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\left(đpcm\right)\)

Không tính kết quả

Bài 2: 

b) Gọi \(d\inƯC\left(21n+4;14n+3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}21n+4⋮d\\14n+3⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\)

\(\Leftrightarrow1⋮d\)

\(\Leftrightarrow d\inƯ\left(1\right)\)

\(\Leftrightarrow d\in\left\{1;-1\right\}\)

\(\LeftrightarrowƯCLN\left(21n+4;14n+3\right)=1\)

hay \(\dfrac{21n+4}{14n+3}\) là phân số tối giản(đpcm)

Bài 1: 

a) Ta có: \(A=1+2-3-4+5+6-7-8+...-299-300+301+302\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+301+302\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)+603\)

\(=75\cdot\left(-4\right)+603\)

\(=603-300=303\)

Bài 1: 

c) Ta có: \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Leftrightarrow3B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

\(\Leftrightarrow3B-B=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

\(\Leftrightarrow2B=1-\dfrac{1}{3^{99}}\)

\(\Leftrightarrow B=\dfrac{3^{99}-1}{3^{99}\cdot2}\)

Bài 2: 

a) Vì tổng của hai số là 601 nên trong đó sẽ có 1 số chẵn, 1 số lẻ

mà số nguyên tố chẵn duy nhất là 2

nên số lẻ còn lại là 599(thỏa ĐK)

Vậy: Hai số nguyên tố cần tìm là 2 và 599

b,Gọi ƯCLN(21n+4,14n+3)=d

21n+4⋮d ⇒42n+8⋮d

14n+3⋮d ⇒42n+9⋮d

(42n+9)-(42n+8)⋮d

1⋮d ⇒ƯCLN(21n+4,14n+3)=1

Vậy phân số 21n+4/14n+3 là phân số tối giản

c,xy-2x+5y-12=0

xy-2x+5y-12+2=0+2

xy-2x+5y-10=2

xy-2x+5y-5.2=-2

x.(y-2)+5.(y-2)=2

(y-2).(x+5)=2

Sau đó bạn tự lập bảng 

Thanks bạn Đinh Tuấn Việt nhiều nah!!!!