\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{\sqrt{4-x}+4}{\sqrt{4-x}+5}\)
\(\Leftrightarrow\dfrac{x}{x+1}=\dfrac{\sqrt{4-x}+4}{\sqrt{4-x}+5}\)
=>\(\sqrt{4-x}+4=x\)
=>\(\sqrt{4-x}=x-4\)
=>x=4
`1/[1.2]+1/[2.3]+...+1/[x(x+1)]=[\sqrt{4-x}+4]/[\sqrt{4-x}+5]` `ĐK: x <= 4`
`<=>1-1/2+1/2-1/3+...+1/x-1/[x+1]=[\sqrt{4-x}+5-1]/[\sqrt{4-x}+5]`
`<=>1-1/[x+1]=1-1/[\sqrt{4-x}+5]`
`<=>1/[x+1]=1/[\sqrt{4-x}+5]`
`<=>\sqr{4-x}+5=x+1`
`<=>4-x+\sqrt{4-x}=0`
`<=>\sqrt{4-x}(\sqrt{4-x}+1)=0`
Mà `\sqrt{4-x}+1 > 0 AA x <= 4`
`=>\sqrt{4-x}=0`
`<=>4-x=0`
`<=>x=4`
