Lời giải:

\(\text{VT}=\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+....+\frac{50-49}{49.50}\)

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

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-\left(1+\frac{1}{2}+\frac{1}{3}....+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{49}+\frac{1}{50}\)

Do đó ta có đpcm.

\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =1+\dfrac{1}{2}+...+\dfrac{1}{50}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{25}\\ =\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)

tại sao \(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(LINH_1=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+....+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+....+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}=LINH_2\left(đpcm\right)\)

Ta có :

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

= (1+ 1/3 + 1/5 + ... + 1/49) - (1/2 + 1/4 + ... +1/50)

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

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

=1/26 + 1/27 +1/28 +...+1/50 (đpcm)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=1-\dfrac{1}{50}\)

\(A=\dfrac{49}{50}\)

Ta có :

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...........+\dfrac{1}{49.50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{50}\)

\(A=\dfrac{49}{50}\)

~ Chúc bn học tốt ~

\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)

Thanks

Đề bài sai vì không có qui luật !

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy \(x=-2004\)