\(2x-\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{12}-...-\dfrac{1}{49\cdot50}=7-\dfrac{1}{50}+x\)

\(\Leftrightarrow2x-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=x+\dfrac{349}{50}\)

\(\Leftrightarrow2x-\dfrac{49}{50}-x-\dfrac{349}{50}=0\)

=>x=398/50=199/25

2x - \(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-....-\frac{1}{49.50}\)= 7-\(\frac{1}{50}\)+x

2x - x - \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\right)\)= 7 - \(\frac{1}{50}\)

x - \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)= \(\frac{349}{50}\)

x - \(\left(1-\frac{1}{50}\right)\)=\(\frac{349}{50}\)

x - \(\frac{49}{50}\)=\(\frac{349}{50}\)

x = \(\frac{349}{50}+\frac{49}{50}\)

x = \(\frac{199}{25}\)

\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-....-\frac{1}{49.50}=7+\frac{1}{50}+x\)

\(2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{49.50}\right)=7+\frac{1}{50}+x\)

\(2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\right)=7+\frac{1}{50}+x\)

\(2x-\left(\frac{1}{1}-\frac{1}{50}\right)=7+\frac{1}{50}+x\)

\(2x-1+\frac{1}{50}=7+\frac{1}{50}+x\)

=> 2x - 1 = 7 + x

=> 2x - x = 7 + 1

=> x = 8 

2x - \(\dfrac{1}{2}-\dfrac{1}{6}-...-\dfrac{1}{49.50}\)= 6-\(\dfrac{1}{50}\) + x

<=> x - ( \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)) = \(\dfrac{299}{50}\)

<=> x - \(\left(1-\dfrac{1}{50}\right)\) = \(\dfrac{299}{50}\)

<=> x - \(\dfrac{49}{50}\) = \(\dfrac{299}{50}\)

<=> x = \(\dfrac{174}{25}\)

1/1.2+1/3.4+1/5.6+...+1/49.50

=1/1-1/2+1/3-1/4+...+1/49-1/50

=1/1+1/2+1/3+1/4+...+1/49+1/50-2(1/2+1/4+1/6+...+1/50)

=1/1+1/2+1/3+1/4+...+1/49+1/50-(1/1+1/2+1/3+1/4+...+1/25)

=1/26+1/27+...+1/50=1/26+1/27+...+1/50(đpcm)

b. 1/1-1/2+1/3-1/4+...+1/99-1/100=99/100

7/12=175/300; 5/6=10/12=250/300; 99/100=297/300

(hình như khúc này đề bài sai hả bạn) bạn tự tính ra nhé

bài 2: a.x+1/10+x/12+x/14+...x+1/20

(x+x+x...+x)+(1/10+1/12+...+1/20)

ko có kết quả sao tìm x được bạn:[

b.x+1/2000+x+2/1999=x+3/1998+x+4/1997

x+1/2000+x+2/1999=x+3/1998+x+4/1997

(x+1/2000+1)+(x+2/1999+1)=(x+3/1998+1)+(x+4/1997+1)

x+2002/2000+x+2002/1999=x+2002/1998+x+2002/1997

x+2002(1/2000+1/1999)=(x+2002)(1/1998+1/1997)

=>(1/2000+1/1999)=(1/1998+1/1997)

x+2002(1/2000+1/1999)-(x+2002)(1/1998+1/1997)=0

(x+2002)(1/2000+1/1999-1/1998-1/1997)=0

(x+2002).0=0

(x+2002)=0

x =0-2002=-2002

Chúc bạn học tốt.

\(2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\) =\(\frac{349}{50}+x\)

\(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\) \(=\frac{349}{50}\)

\(x-\left(1-\frac{1}{50}\right)=\frac{349}{50}\)

\(x-\frac{49}{50}=\frac{349}{50}\)

\(x=\frac{199}{25}\)

=> 2x- ( 1/2+1/6+1/12+..._1/ 49.50 )= 7-1/50+x

=> 2x -( 1/1.2 + 1/2.3+1/3.4+...+1/49.50)= 7-1/50+x

=> 2x - ( 1- 1/2+ 1/2-1/3+1/3-1/4+...+1/49-1/50) = 7-1/50 + x

=> 2x - ( 1-1/50) =7-1/50 + x

=> 2x- 1+ 1/50=7-1/50+ x

=> 1+1/50= 2x- (7 - 1/50+ x)

=> 1+1/50 = 2x- 7 + 1/50- x

=> 1+1/50 = x + 1/50 - 7

=> 1 = x + 1/50 - 7 - 1/50

=> 1 = x - 7

=> x = 7+ 1

=> x = 8

\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=7-\frac{1}{50}+x\)

\(\Leftrightarrow2x-\left[\frac{1}{2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\right]=7-\frac{1}{50}+x\)

\(\Leftrightarrow2x-\left[\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\right]=7-\frac{1}{50}+x\)

\(\Leftrightarrow2x-\left[\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{50}\right)\right]=7-\frac{1}{50}+x\)

\(\Leftrightarrow2x-\left[\frac{1}{2}-\frac{12}{50}\right]=7-\frac{1}{50}+x\)

\(\Leftrightarrow2x-\frac{1}{50}=\frac{349}{50}+x\)

\(\Leftrightarrow2x-x=\frac{349}{50}+\frac{1}{50}\)

\(\Rightarrow x=7\)

Lời giải:
\(A=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-998}{999}.\frac{-999}{1000}\\ =\frac{(-1)(-2)(-3)...(-998)(-999)}{2.3.4....1000}\\ =-\frac{1.2.3.4....998.999}{2.3.4...1000}\\ =-\frac{1}{1000}\)

Trong $B$ có một thừa số là $1-\frac{7}{7}=0$ nên $B=0$ (do số nào nhân với $0$ cũng sẽ bằng $0$.

----------------------

$C=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{49.51}{50^2}$

$=\frac{1.3.2.4.3.5.....49.51}{2^2.3^2.4^2....50^2}$

$=\frac{(1.2.3...49)(3.4.5...51)}{(2.3.4...50)(2.3.4...50)}$
$=\frac{1.2.3...49}{2.3.4...50}.\frac{3.4.5...51}{2.3.4....50}$

$=\frac{1}{50}.\frac{51}{2}=\frac{51}{100}$

$D=\frac{2^2.3^2.4^2....50^2}{1.3.2.4.3.5....49.50}$

$=\frac{(2.3.4...50)(2.3.4...50)}{(1.2.3...49)(3.4....50)}$
$=\frac{2.3.4...50}{1.2.3...49}.\frac{2.3.4....50}{3.4....50}$
$=50.2=100$