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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 : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.....+\frac{1}{50}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-....-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+.......+\frac{1}{50}\)
giúp mik vs, mik bik các pạn giờ này đang ngủ rùi nhưng giúp mik lần này thui.yêu các pạn nhìu
\(5\frac{1}{2}+\left(-3\right)=\frac{11}{2}+\frac{-3}{1}\)\(=\frac{11}{2}+\frac{-6}{2}=\frac{5}{2}\)\(;\)
\(4\frac{9}{11}+\left(-2\frac{1}{11}\right)=\frac{53}{11}+\frac{-23}{11}\)\(=\frac{30}{11}\)\(;\)
\(2\frac{1}{2}+\left(-6\right)=\frac{5}{2}+\frac{-6}{1}\)\(=\frac{5}{2}+\frac{-12}{2}=\frac{-7}{2}\)\(;\)
\(\left(-\frac{4}{5}\right)+\frac{1}{2}=\frac{-4}{5}+\frac{1}{2}\)\(=\frac{-8}{10}+\frac{5}{10}=\frac{-3}{10}\)\(;\)
\(4,3-\left(-1,2\right)=4,3+1,2=5,5\)\(=\frac{55}{10}=\frac{11}{2}\)\(;\)
\(0-\left(-0,4\right)=0+0,4=0,4\)\(=\frac{4}{10}=\frac{2}{5}\)\(;\)
\(\frac{-2}{3}-\frac{-1}{3}=\frac{-2}{3}+\frac{1}{3}=\frac{-1}{3}\)\(;\)
\(\frac{-1}{2}-\frac{-1}{6}=\frac{-1}{2}+\frac{1}{6}\)\(=\frac{-3}{6}+\frac{1}{6}=\frac{-2}{6}=\frac{-1}{3}\)\(;\)
\(x+\frac{1}{3}=\frac{3}{4}\) \(;\) \(x-\frac{2}{5}=\frac{5}{7}\) \(;\)
\(x=\frac{3}{4}-\frac{1}{3}\) \(x=\frac{5}{7}+\frac{2}{5}\)
\(x=\frac{5}{12}\) \(x=\frac{39}{35}\)
\(-x-\frac{2}{3}=-\frac{6}{7}\) \(;\) \(\frac{4}{7}-x=\frac{1}{3}\)
\(\frac{6}{7}-\frac{2}{3}=x\) \(\frac{4}{7}-\frac{1}{3}=x\)
\(\frac{4}{21}=x\) \(\Leftrightarrow\)\(x=\frac{4}{21}\) \(\frac{5}{21}=x\)\(\Leftrightarrow\)\(x=\frac{5}{12}\)
Câu a đề thiếu vế phải rồi bạn
b: \(\Leftrightarrow x\cdot0+1=0\)
=>0x+1=0(vô lý)
Ta có :
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
a: Đề thiếu vế phải rồi bạn
b: \(\Leftrightarrow x\cdot\dfrac{-12}{13}=6+\dfrac{1}{13}+5=11+\dfrac{1}{13}=\dfrac{144}{13}\)
hay x=-12
c: \(\Leftrightarrow x\cdot\dfrac{-5}{4}-\dfrac{6}{5}x=\dfrac{-1}{2}-\dfrac{3}{7}\)
\(\Leftrightarrow x\cdot\dfrac{-49}{20}=\dfrac{-13}{14}\)
hay x=130/343
Ta có 4A=\(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)
Trừ 4A cho A ta được
3A = \(1-\frac{1}{2^{100}}\)=> 3A <1 => A<1/3 (đpcm)
Chúc bạn học tốt
Ta có :\(A=\frac{1}{2^2}+...+\frac{1}{2^{100}}\)
\(2A=\frac{1}{2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^{100}}\)
Lại có :
\(\frac{1}{3}=\frac{1}{2}-\frac{1}{6}\)
Vì \(\frac{1}{2^{100}}< \frac{1}{6}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2^{100}}>\frac{1}{2}-\frac{1}{6}\)
\(\Rightarrow A>\frac{1}{3}\)
Vậy \(A>\frac{1}{3}\)(ĐPCM)
a) \(\frac{2}{7}x-\frac{1}{3}x=\frac{5}{21}\)
\(\left(\frac{2}{7}-\frac{1}{3}\right)x=\frac{5}{21}\)
\(\left(-\frac{1}{21}\right)x=\frac{5}{21}\Rightarrow x=\frac{5}{21}:-\frac{1}{21}=-5\)
b) \(\frac{x+1}{1974}+\frac{x+2}{1973}+\frac{x+3}{1972}=-3\)
\(\left(\frac{x+1}{1974}+1\right)+\left(\frac{x+2}{1973}+1\right)+\left(\frac{x+3}{1972}+1\right)=-3+3\)
\(\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)
\(\left(x+1975\right)\left(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\right)=0\)
Mà \(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}>0\Rightarrow x+1975=0\)
\(x=-1975\)
\(2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\) =\(\frac{349}{50}+x\)
\(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\) \(=\frac{349}{50}\)
\(x-\left(1-\frac{1}{50}\right)=\frac{349}{50}\)
\(x-\frac{49}{50}=\frac{349}{50}\)
\(x=\frac{199}{25}\)
=> 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