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

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

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

=>x=398/50=199/25

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

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

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

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

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

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

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

\(\Rightarrow2x-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{49.50}\right)=7-\dfrac{1}{50}+x\)

\(\Rightarrow2x-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=7-\dfrac{1}{50}+x\)

\(\Rightarrow2x-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=7-\dfrac{1}{50}+x\)\(\Rightarrow2x-1+\dfrac{1}{50}=7-\dfrac{1}{50}+x\)

\(\Rightarrow2x=7-\dfrac{1}{50}+x-\dfrac{1}{50}+1\)

\(\Rightarrow2x=\dfrac{199}{25}+x\)

\(\Rightarrow x=\dfrac{199}{25}\)

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

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

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

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

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

=> 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

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

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

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

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

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

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

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

\(\Rightarrow x=7\)

1.Tìm x , biết

.2x -1/2-1/6-1/12-...- 1/49*50=7-1/50+x

=> 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 = 8 

Vậy...

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\(\text{2x -1/2-1/6-1/12-...- 1/49*50=7-1/50+x }\)

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

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

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

=> \(\text{2x - ( 1-1/50) =7-1/50 + x }\)

=>\(\text{ 2x- 1+ 1/50=7-1/50+ x }\)

=> \(\text{1+1/50= 2x- (7 - 1/50+ x) }\)

=>\(\text{ 1+1/50 = 2x- 7 + 1/50- x}\)

=>\(\text{ 1+1/50 = x + 1/50 - 7 }\)

=>\(\text{ 1 = x + 1/50 - 7 - 1/50 }\)

=> \(\text{1 = x - 7}\)

=> \(\text{x = 8 }\)