Question

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

b) \(\frac{x+4}{200}+\frac{x+3}{201}=\frac{x+2}{202}+\frac{x+1}{203}\)

Answer 1

\(a;\)

\(=0-1\)

\(=-1\)

Answer 2

\(b;\)

\(=0-4\)

\(=-4\)

Answer 3

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{15}\)

\(\frac{181\left(x+1\right)}{660}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{181\left(x+1\right)}{660}=\frac{17\left(x+1\right)}{52}\)

\(2353\left(x+1\right)=2805\left(x+1\right)\)

\(2353x+2353=2805x+2805\)

\(2353=2805x+2805-2353x\)

\(2353=452x+2805\)

\(2353-2805=452x\)

\(-452=452x\)

\(x=-1\)

Answer 4

b) \(\frac{x+4}{200}+\frac{x+3}{201}=\frac{x+2}{202}+\frac{x+1}{203}\)

\(\Leftrightarrow\frac{x+4}{200}+1+\frac{x+3}{201}+1=\frac{x+2}{202}+1+\frac{x+1}{203}+1\)

\(\Leftrightarrow\frac{x+204}{200}+\frac{x+204}{201}=\frac{x+204}{202}+\frac{x+204}{203}\)

\(\Leftrightarrow\frac{x+204}{200}+\frac{x+204}{201}-\frac{x+204}{202}-\frac{x+204}{203}=0\)

\(\Leftrightarrow\left(x+204\right)\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)

Mà \(\left(\frac{1}{200}>\frac{1}{201}>\frac{1}{202}>\frac{1}{203}\right)\)nên \(\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)\ne0\)

\(\Rightarrow x+204=0\Leftrightarrow x=-204\)