Question

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tìm x

1, \(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

2, \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)

3, \(x-\left(\frac{11}{12}-x\right)=x-\frac{3}{4}\)

4, \(-29-4\cdot|3x+6|=-41\)

5, \(\frac{1}{5}\cdot2x+\frac{1}{3}\cdot2^{x+1}=\frac{1}{5}\cdot2^7+\frac{1}{3}\cdot2^8\)

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Answer 1

[Kiệt Nguyễn]

\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x=0+\frac{2}{5}\)

\(\Leftrightarrow x\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)

\(\Leftrightarrow x\left(\frac{5}{15}+\frac{6}{15}\right)=\frac{2}{5}\)

\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)

\(\Leftrightarrow x=\frac{2}{5}\div\frac{11}{15}\)

\(\Leftrightarrow x=\frac{6}{11}\)

Answer 2

[Kiệt Nguyễn]

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{49}{50}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{49}{50}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{49}{50}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{50}\div2\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{50}\times\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{49}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{50}{100}-\frac{49}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)

\(\Leftrightarrow x+1=100\)

\(\Leftrightarrow x=100-1\)

\(\Leftrightarrow x=99\)