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\(\left(\frac{3}{1.3}+\frac{3}{3.5}+.......+\frac{3}{97.99}\right).\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow[\frac{3}{2}.(\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{2}{97.99})].\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow[\frac{3}{2}.(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{97}-\frac{1}{99})].\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow[\frac{3}{2}.(1-\frac{1}{99})].\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow\left(\frac{3}{2}.\frac{98}{99}\right).\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow\frac{49}{33}.\left(2x+1\right)=x+\frac{1}{33}\)
\(\Rightarrow\frac{49}{33}.2x+\frac{49}{33}=x+\frac{1}{33}\)
\(\Rightarrow\frac{98}{33}.x+\frac{49}{33}=x+\frac{1}{33}\)
\(\Rightarrow\frac{98}{33}.x-x=\frac{1}{33}-\frac{49}{33}\)
\(\Rightarrow\frac{65}{33}.x=\frac{-16}{11}\)
\(\Rightarrow x=\frac{-16}{11}:\frac{65}{33}\)
\(\Rightarrow x=\frac{-48}{65}\)
Vậy \(x=\frac{-48}{65}\)
VT = 1/2.( 1-1/3+1/3-1/5+...+ 2/49-1/51)
= 1/2. 50/51
=> 6x-5/10+10 = 25/51
............. Tụ làm phàn còn lại nhé
Nhân cả 2 vê với 2 ta được:
\(\frac{2.\left(6x-5\right)}{20}\)=\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+...+\(\frac{2}{49.51}\)
<=>\(\frac{6x-5}{10}\)=\(1-\frac{1}{3}+\)\(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\)
<=>\(\frac{6x-5}{10}=1-\frac{1}{51}\)
<=>\(6x-5=\frac{50}{51}.10\)
<=>\(x=\frac{755}{306}\)
để ý 1+1/x(x+2)=(x2+2x+1)/x(x+2)=(x+1)2/x(x+2)
+ 1+1/1.3=22/1.3 ;......
1. \(\frac{3x-7}{5}=\frac{2x-1}{3}\)
<=> 3(3x-7)=5(2x-1)
<=> 9x-21=10x-5
<=> -21+5=10x-9x
<=> x=-16
2. \(\frac{3x-7}{2}+\frac{2x-1}{3}=-16\)
<=> \(\frac{3\left(3x-7\right)}{6}+\frac{2\left(2x-1\right)}{6}=\frac{-96}{6}\)
=> 9x-21+4x-2=-96
<=> 13x-23=-96
<=> 13x=-73
<=> x=\(\frac{-73}{13}\)
3. \(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
<=> \(\frac{15x}{15}-\frac{5\left(x+1\right)}{15}=\frac{3\left(2x+1\right)}{15}\)
=> 15x-5x-5=6x+3
<=> 15x-5x-6x=3+5
<=> 4x=8
<=> x=2
4. \(\frac{7-3x}{12}+\frac{3}{4}=2\left(x-2\right)+\frac{5-\left(5-2x\right)}{6}\)
<=>\(\frac{7-3x}{12}+\frac{9}{12}=\frac{24\left(x-2\right)}{12}+\frac{2\left[5-\left(5-2x\right)\right]}{12}\)
=> 7-3x+9=24x-48+4x
<=> -3x-24x-4x=-48-7
<=> -31x=-55
<=> x= \(\frac{55}{31}\)
5. \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
<=> \(\frac{7\left(2x-1\right)}{21}-\frac{3\left(5x+2\right)}{21}=\frac{21\left(x+13\right)}{21}\)
=> 14x-7-15x-6=21x+273
<=> 14x-15x-21x=273+7+6
<=> -22x=286
<=> x= -13
a/\(\Leftrightarrow3\left(3x-7\right)=5\left(2x-1\right)\Leftrightarrow9x-21=10x-5\Leftrightarrow x=-16\)
b/\(\Leftrightarrow\frac{9x-21+4x-2}{6}=-16\)\(\Leftrightarrow13x-23=-96\Leftrightarrow x=x=-\frac{73}{13}\)
c/\(\Leftrightarrow\frac{3x-x+1}{3}-\frac{2x+1}{5}=0\Leftrightarrow\left(2x+1\right)\left(\frac{1}{3}-\frac{1}{5}\right)=0\Leftrightarrow x=-\frac{1}{2}\)
a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)
\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)
=>-9/10=-9/10(luôn đúng)
b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)
=>347x+780=1552
=>347x=772
hay x=772/347
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+....+\frac{1}{47.48.49.50}\)
\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}.\frac{6533}{39200}=\frac{6533}{117600}\)
Ta có: \(\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\right)\cdot\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\cdot\left(x-1\right)=\frac{3x}{5}-\frac{7}{15}\)
\(\Leftrightarrow\frac{14}{15}\cdot\left(x-1\right)=\frac{9x-7}{15}\)
\(\Leftrightarrow x-1=\frac{9x-7}{15}:\frac{14}{15}=\frac{9x-7}{14}\)
hay \(x=\frac{9x-7}{14}+1=\frac{9x-7}{14}+\frac{14}{14}=\frac{9x+7}{14}\)
\(\Leftrightarrow x\cdot14=9x+7\)
\(\Leftrightarrow14x-9x-7=0\)
\(\Leftrightarrow5x-7=0\)
\(\Leftrightarrow5x=7\)
hay \(x=\frac{7}{5}\)
Vậy: \(x=\frac{7}{5}\)