\(B=\dfrac{49}{2\cdot9}+\dfrac{49}{9\cdot16}+\dfrac{49}{16\cdot23}+...+\dfrac{49}{65\cdot72}\)

\(B=\dfrac{49}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)

\(B=7\cdot\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)

\(B=7\cdot\left(\dfrac{36}{72}-\dfrac{1}{72}\right)\)

\(B=7\cdot\dfrac{35}{72}\)

\(B=\dfrac{\left(7\cdot35\right)}{72}\)

\(B=\dfrac{245}{72}\)

\(\dfrac{B}{7}=\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+\dfrac{7}{16\cdot23}+...+\dfrac{49}{65\cdot72}\\ \dfrac{B}{7}=\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\\ \dfrac{B}{7}=\dfrac{1}{2}-\dfrac{1}{72}\\ \dfrac{B}{7}=\dfrac{35}{72}\\ B=\dfrac{35}{72}\times7\\ B=\dfrac{245}{72} \)

(1/49-1/9)(1/49-1/16).....(1/49-1/49)...(1/49-1/2401)

(1/49-1/9)(1/49-1/16).....0....(1/49-1/2401)

=0

vì khoảng cách là 7 mà 49 chia hết cho 7 nên có phân số 1/49 trong dãy

1.name

2.My

3.is

4.book

5.color

6.blue

1/ What is your name ?

2/ My name is Andy.

3/ What is this?

4/ It’s a book.

5/ What color is this?

6/ It’s a blue pencil.

1/name

2/my

3/is

4/book

5/blue

hơi ngu tí nếu sai mong mn sửa hộ

ĐKXĐ : x khác 49 , x khác 50

Ta có :

\(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)

\(\Leftrightarrow\frac{x-49}{50}-1+\frac{x-50}{49}-1=\frac{49}{x-50}-1+\frac{50}{x-49}-1\)

\(\Leftrightarrow\frac{x-99}{50}+\frac{x-99}{49}=\frac{99-x}{x-50}+\frac{99-x}{x-49}\)

\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{x-50}+\frac{1}{x-49}\right)=0\) 

Mà 1/50 + 1/49 + 1/x-50 + 1/x-49 khác 0

\(\Leftrightarrow x-99=0\)

\(\Leftrightarrow x=99\)

@Kyo-kun

\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{50}+\frac{1}{x-50}+\frac{1}{49}+\frac{1}{x-49}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-99=0\\\frac{x}{50\left(x-50\right)}+\frac{x}{49\left(x-49\right)}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=99\\x\left(\frac{1}{50\left(x-50\right)}+\frac{1}{49\left(x-49\right)}\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=99\\x=0\end{cases}\left(t.m\right)}}\)

Vậy x = 99 hoặc x = 0

\(F=\dfrac{49}{2.9}+\dfrac{49}{9.16}+............+\dfrac{49}{65.72}\)

\(\Leftrightarrow F=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+............+\dfrac{7^2}{65.72}\)

\(\Leftrightarrow F=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+.............+\dfrac{7}{65.72}\right)\)

\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...........+\dfrac{1}{65}-\dfrac{1}{75}\right)\)

\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)

\(\Leftrightarrow F=7.\dfrac{35}{72}=\dfrac{245}{72}\)

\(G=\dfrac{3}{1.3}+\dfrac{3}{3.5}+...........+\dfrac{3}{47.49}\)

\(\Leftrightarrow G=\dfrac{3.2}{1.3.2}+\dfrac{3.2}{3.5.2}+........+\dfrac{3.2}{47.49}\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+..........+\dfrac{2}{47.49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}.\dfrac{48}{49}=\dfrac{72}{49}\)

-49 - 49 x 49 = -49 - 2401

                    =-2450

\(\left(-49\right)-49\cdot49=\left(-49\right)-\left[49\cdot49\right]=\left(-49\right)-2401=-2450\)

k mình

k lại

47+145+124+100

=416 : 49

=416/49