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\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
\(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Rightarrow\frac{14}{5}x-15=34\)
\(\frac{\Rightarrow14}{5}x=49\)
\(\Rightarrow x=\frac{35}{2}\)
\(3\frac{2}{7}x-\frac{1}{8}=2\frac{3}{4}\)
\(\Rightarrow\frac{23}{7}x-\frac{1}{8}=\frac{11}{4}\)
\(\frac{\Rightarrow23}{7}x=\frac{23}{8}\)
\(\Rightarrow x=\frac{7}{8}\)
nhanh thế anh bạn định giải lun mà nhanh quá
a) 51 1/5 . 3/8 + 27 1/5 . -3/8 + 19
= ( 51 1/5 + 27 1/5 ) . -1 + 19
= ( 51 + 27 1/5 ) . -1 + 19
= 78 1/5 . -1 + 19
= 391/5 . -1 + 19
= -391/5 + 19
= -106/15
b) 1/6 . ( -2 3/2 ) + 1 2/3 . ( -13/5 )
= 1/6 . -13/5 + 5/3 . -13/5
= 1/6 . 5/3 . ( -13/5 + -13/5 )
= 1/6 . 5/3 . -26/5
= 1.5.-26 / 6 . 3 . 5
= -13/30
2: =>2x-1/4=5/6-1/2x
=>5/2x=5/6+1/4=13/12
=>x=13/30
3: =>3x-5/6=2/3-1/2x
=>3,5x=2/3+5/6=4/6+5/6=9/6=3,2
hay x=32/35
e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)
\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)
\(=\frac{1}{7}.\left(-2\right)\)
\(=-\frac{2}{7}.\)
Chúc bạn học tốt!
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow1-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{50}\)
\(\Rightarrow n+1=50\)
\(\Rightarrow n=49\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{2n+1}=\frac{1}{51}\)
\(\Rightarrow2n+1=51\)
\(\Rightarrow2n=50\)
\(\Rightarrow n=25\)
\(A=-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{51}}-\frac{1}{3^{52}}\)
\(\hept{\begin{cases}A=B.\frac{1}{3}\\A+B=-\frac{1}{3}-\frac{1}{3^{52}}\end{cases}}\)
Giải hệ phương trình Bậc nhất 2 ẩn trên Bằng Phương pháp trừ Đại số
ta được \(B=\frac{\left(-1-\frac{1}{3^{51}}\right)}{4}=\frac{-\left(3^{51}+1\right)}{4.3^{51}}\)