Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-....-\frac{1}{49.50}=7+\frac{1}{50}+x\)
\(2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{49.50}\right)=7+\frac{1}{50}+x\)
\(2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\right)=7+\frac{1}{50}+x\)
\(2x-\left(\frac{1}{1}-\frac{1}{50}\right)=7+\frac{1}{50}+x\)
\(2x-1+\frac{1}{50}=7+\frac{1}{50}+x\)
=> 2x - 1 = 7 + x
=> 2x - x = 7 + 1
=> x = 8
Gửi link thì bị lỗi, thôi nhai lại v:
Xét VT__Ta có: \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+..+\frac{1}{50}\right)\)
\(=\) \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{50}-1+\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.....+\frac{1}{50}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-....-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+.......+\frac{1}{50}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{60}\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+...+\frac{1}{50}+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
2/ \(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=\frac{7}{12}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}>\frac{7}{12}\)
Tương tự câu trên ta có: \(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(A=\frac{1}{51}+...+\frac{1}{60}+\frac{1}{61}+...+\frac{1}{70}+\frac{1}{71}+...+\frac{1}{80}+\frac{1}{81}+...+\frac{1}{90}+\frac{1}{91}+...+\frac{1}{100}\)
\(A< \frac{1}{50}+...+\frac{1}{50}+\frac{1}{60}+...+\frac{1}{60}+\frac{1}{70}+...+\frac{1}{70}+\frac{1}{80}+...+\frac{1}{80}+\frac{1}{90}+...+\frac{1}{90}\)
\(A< 10.\frac{1}{50}+10.\frac{1}{60}+10.\frac{1}{70}+10.\frac{1}{80}+10.\frac{1}{90}\)
\(A< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{5}{6}\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
Khử mẫu : \(12x-12+6x-6=4x+3x-7\)
\(\Leftrightarrow18x-18=7x-7\Leftrightarrow11x=11\Leftrightarrow x=1\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12+6x-6}{12}=\frac{4x+3x-7}{12}\)
\(\Leftrightarrow18x-18=7x-7\)
\(\Leftrightarrow18x+7x=18+7\)
\(\Leftrightarrow25x=25\)
\(\Leftrightarrow x=1\)
giúp mik vs, mik bik các pạn giờ này đang ngủ rùi nhưng giúp mik lần này thui.yêu các pạn nhìu
\(5\frac{1}{2}+\left(-3\right)=\frac{11}{2}+\frac{-3}{1}\)\(=\frac{11}{2}+\frac{-6}{2}=\frac{5}{2}\)\(;\)
\(4\frac{9}{11}+\left(-2\frac{1}{11}\right)=\frac{53}{11}+\frac{-23}{11}\)\(=\frac{30}{11}\)\(;\)
\(2\frac{1}{2}+\left(-6\right)=\frac{5}{2}+\frac{-6}{1}\)\(=\frac{5}{2}+\frac{-12}{2}=\frac{-7}{2}\)\(;\)
\(\left(-\frac{4}{5}\right)+\frac{1}{2}=\frac{-4}{5}+\frac{1}{2}\)\(=\frac{-8}{10}+\frac{5}{10}=\frac{-3}{10}\)\(;\)
\(4,3-\left(-1,2\right)=4,3+1,2=5,5\)\(=\frac{55}{10}=\frac{11}{2}\)\(;\)
\(0-\left(-0,4\right)=0+0,4=0,4\)\(=\frac{4}{10}=\frac{2}{5}\)\(;\)
\(\frac{-2}{3}-\frac{-1}{3}=\frac{-2}{3}+\frac{1}{3}=\frac{-1}{3}\)\(;\)
\(\frac{-1}{2}-\frac{-1}{6}=\frac{-1}{2}+\frac{1}{6}\)\(=\frac{-3}{6}+\frac{1}{6}=\frac{-2}{6}=\frac{-1}{3}\)\(;\)
\(x+\frac{1}{3}=\frac{3}{4}\) \(;\) \(x-\frac{2}{5}=\frac{5}{7}\) \(;\)
\(x=\frac{3}{4}-\frac{1}{3}\) \(x=\frac{5}{7}+\frac{2}{5}\)
\(x=\frac{5}{12}\) \(x=\frac{39}{35}\)
\(-x-\frac{2}{3}=-\frac{6}{7}\) \(;\) \(\frac{4}{7}-x=\frac{1}{3}\)
\(\frac{6}{7}-\frac{2}{3}=x\) \(\frac{4}{7}-\frac{1}{3}=x\)
\(\frac{4}{21}=x\) \(\Leftrightarrow\)\(x=\frac{4}{21}\) \(\frac{5}{21}=x\)\(\Leftrightarrow\)\(x=\frac{5}{12}\)
ak xin lỗi mk ghi lộn đề 
, đề đúng là:
Chứng minh rằng: \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Các bạn giúp mk với mk cần gấp thank you!!! 
#)Giải :
\(2x-3=x+\frac{1}{2}\)
\(\Leftrightarrow2x-3-x+\frac{1}{2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\x=-\frac{1}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}}\)
a) \(2x-3=x+\frac{1}{2}\)
\(\Leftrightarrow2x-x=\frac{1}{2}+3\)
\(\Leftrightarrow x=\frac{7}{2}\)
Vậy...
b) \(4x-\left(2x+1\right)=3-\frac{1}{3}+x\)
\(\Leftrightarrow4x-2x-1=3-\frac{1}{3}+x\)
\(\Leftrightarrow4x-2x-x=3-\frac{1}{3}+1\)
\(\Leftrightarrow x=\frac{11}{3}\)
Vậy ...
c) \(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=7-\frac{1}{50}+x\)
\(\Leftrightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(1-\frac{1}{50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\frac{49}{50}=\frac{349}{50}+x\)
\(\Leftrightarrow2x-x=\frac{349}{50}+\frac{49}{50}\)
\(\Leftrightarrow x=\frac{199}{25}\)
Vậy ...
\(2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\) =\(\frac{349}{50}+x\)
\(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\) \(=\frac{349}{50}\)
\(x-\left(1-\frac{1}{50}\right)=\frac{349}{50}\)
\(x-\frac{49}{50}=\frac{349}{50}\)
\(x=\frac{199}{25}\)
=> 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