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cho thêm điều kiện x,y thuộc Z nữa nhá
\(\frac{3}{x}+\frac{1}{3}=\frac{y}{3}\)
\(\frac{3}{x}=\frac{y-1}{3}\)
\(\Rightarrow x.\left(y-1\right)=9\)
Lập bảng ta có :
| x | 1 | 9 | -1 | -9 | 3 | -3 |
| y-1 | 9 | 1 | -9 | -1 | 3 | -3 |
| y | 10 | 2 | -8 | 0 | 4 | -2 |
Vậy ( x ; y ) = { ( 1 ; 10 ) ; ( 9 ; 2 ) ; ( -1 ; -8 ) ; ( -9 ; 0 ) ; ( 3 ; 4 ) ; ( -3 ; -2 ) }
mấy bài còn lại làm tương tự
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow1-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{50}\)
\(\Rightarrow n+1=50\)
\(\Rightarrow n=49\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{2n+1}=\frac{1}{51}\)
\(\Rightarrow2n+1=51\)
\(\Rightarrow2n=50\)
\(\Rightarrow n=25\)
Đặt :
\(A=\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\)
\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)
\(\Leftrightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)
\(\Leftrightarrow2A=1-\frac{1}{3^{99}}< 1\)
\(\Leftrightarrow A< \frac{1}{2}\left(đpcm\right)\)
Đặt \(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\)
\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Rightarrow2C=1-\frac{1}{3^{99}}< 1\)
=> C = (1 - 1/399)/2 < 1/2
Vậy 1/3 + 1/32 + 1/33 + ....+ 1/399 < 1/2
\(c)\)
\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=\left(7-\frac{1}{50}+x\right)\)
\(\Rightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\left(\frac{350}{50}-\frac{1}{50}+x\right)\)
\(\Rightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Rightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)-x=\frac{349}{50}\)
\(\Rightarrow x-\left(1-\frac{1}{50}\right)=\frac{349}{50}\)
\(\Rightarrow x-\frac{49}{50}=\frac{349}{50}\)
\(\Rightarrow x=\frac{349}{50}+\frac{49}{50}\)
\(\Rightarrow x=\frac{199}{25}\)
Vậy \(x=\frac{199}{25}\)
~ Ủng hộ nhé
\(a)2.x-3=x+\frac{1}{2}\)
\(\Rightarrow2x-3-x=\frac{1}{2}\)
\(\Rightarrow x-3=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+3\)
\(\Rightarrow x=\frac{1}{2}+\frac{6}{2}\)
\(\Rightarrow x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}\)
\(b)4.x-\left(2.x+1\right)=3-\frac{1}{3}+x\)
\(\Rightarrow4.x-2.x-1=\frac{9}{3}-\frac{1}{3}+x\)
\(\Rightarrow2.x-1=\frac{8}{3}+x\)
\(\Rightarrow2x-1-x=\frac{8}{3}\)
\(\Rightarrow x-1=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}+1\)
\(\Rightarrow x=\frac{8}{3}+\frac{3}{3}\)
\(\Rightarrow x=\frac{11}{3}\)
Vậy \(x=\frac{11}{3}\)
~ Ủng hộ nhé
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
\(\Leftrightarrow\frac{13}{36}x=-\frac{8}{45}\)
\(\Rightarrow x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right).\left(-\frac{2}{3}\right)+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{4}{9}x+\frac{1}{3}+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{4}{9}x=\frac{77}{60}\)
\(\Rightarrow x=\frac{231}{80}\)
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
=> \(\frac{4}{9}x-\frac{1}{3}x+\frac{2}{5}-\frac{2}{9}+\frac{1}{4}x=0\)
=> \(\left(\frac{4}{9}x-\frac{1}{3}x+\frac{1}{4}x\right)+\left(\frac{2}{5}-\frac{2}{9}\right)=0\)
=> \(\frac{13}{36}x+\frac{8}{45}=0\)
=> \(\frac{13}{36}x=-\frac{8}{45}\)
=> \(x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}+\frac{1}{5}=\frac{-3}{4}\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}=-\frac{19}{20}\)
=> \(\frac{2}{3}x-\frac{1}{2}=\left(-\frac{19}{20}\right):\left(-\frac{2}{3}\right)=\left(-\frac{19}{20}\right)\cdot\left(-\frac{3}{2}\right)=\frac{57}{40}\)
=> \(\frac{2}{3}x=\frac{57}{40}+\frac{1}{2}=\frac{77}{40}\)
=> \(x=\frac{77}{40}:\frac{2}{3}=\frac{77}{40}\cdot\frac{3}{2}=\frac{231}{80}\)
\(a)\) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)
\(\Leftrightarrow\)\(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+\frac{x+4+96}{96}=0\)
\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)
Nên \(x+100=0\)
\(\Rightarrow\)\(x=-100\)
Vậy \(x=-100\)
Chúc bạn học tốt ~
\(b)\) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=1-\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{2009}\)
\(\Leftrightarrow\)\(x+1=2009\)
\(\Leftrightarrow\)\(x=2009-1\)
\(\Leftrightarrow\)\(x=2008\)
Vậy \(x=2008\)
Chúc bạn học tốt ~
1) \(+2x+3y⋮17\)
\(\Rightarrow26x+39y⋮17\)
\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)
Mà \(17x+34y⋮17\)
\(\Rightarrow9x+5y⋮17\)
\(+9x+5y⋮17\)
\(\Rightarrow36x+20y⋮17\)
\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)
Mà \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)
\(A=1-\frac{1}{2^{99}}\)