Tìm x
\(x+\frac{1}{3}=\frac{4}{3}\)
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(1/1×2 + 1/2×3 + ... + 1/9×10) × x < 2/1×3 + 2/3×5 + ... + 2/9×11
(1 - 1/2 + 1/2 - 1/3 + ... + 1/9 - 1/10) × x < 1 - 1/3 + 1/3 - 1/5 + ... + 1/9 - 1/11
(1 - 1/10) × x < 1 - 1/11
9/10 × x < 10/11
x < 10/11 : 9/10
x < 10/11 × 10/9
x < 100/99
Mà x là số tự nhiên => x = 0 hoặc 1
b) \(\frac{4}{9}x-\frac{1}{2}=\frac{-5}{9}\)
\(\Rightarrow\frac{4}{9}x=\frac{-5}{9}+\frac{1}{2}\)
\(\Rightarrow\frac{4}{9}x=\frac{-1}{18}\)
\(\Rightarrow x=\frac{-1}{18}:\frac{4}{9}\)
\(\Rightarrow x=\frac{-1}{8}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(x=2002\)
Vậy x = 2002
\(\frac{2-x}{x+3}=\frac{6}{5}\)
<=>\(5\left(2-x\right)=6\left(x+3\right)\)
<=>\(10-5x=6x+18\)
<=>\(\left(-5x\right)-6x=18-10\)
<=>\(-11x=8\)
<=>\(x=\frac{-8}{11}\)
\(\frac{5}{3}x-\frac{2}{5}x=\frac{19}{10}\)
\(\left(\frac{5}{3}-\frac{2}{5}\right)x=\frac{19}{10}\)
\(\frac{19}{15}x=\frac{19}{10}\)
\(x=\frac{19}{30}\)
\(\frac{5}{3}x-\frac{2}{5}x=\frac{19}{10}\)
(5/3 - 2/5)x = 19/10
19/15x = 19/10
x = 19/30
\(x+\frac{1}{3}=\frac{4}{3}\)
\(x=\frac{4}{3}-\frac{1}{3}\)
\(x=\frac{3}{3}=1\)
k mình nhaMinh Hiền
x=\(\frac{4}{3}-\frac{1}{3}=1\)