`3)sqrt{1/a^2+1/b^2+1/c^2}`
`=\sqrt{1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)-2/(ab)-2/(bc)-2/(ca)}`
`=\sqrt{(1/a+1/b+1/c)^2-(2(a+b+c))/(abc)}`
`=sqrt{(1/a+1/b+1/c)^2}`
`=|1/a+1/b+1/c|`
4)Khó hơn ta dùng công thức:
`\sqrt{1+1/a^2+1/(a+1)^2}=1+1/a-1/(a+1)(a>0)`
Rất đơn giản:
`VT=sqrt{1+1/a^2+2/a+1/(a+1)^2-2/a}`
`=sqrt{(1+1/a)^2-2/a+1/(a+1)^2}`
`=sqrt{((a+1)/a)^2-2*(a+1)/a*1/(a+1)+1/(a+1)^2}`
`=\sqrt{(1+1/a-1/(a+1))^2}`
`=|1+1/a-1/(a+1)|`
Vì `a>0=>a+1>a`
`=>1/a>1/(a+1)`
`=>1+1/a-1/(a+1)>0`
`=>VT=1+1/a-1/(a+1)(Đpcm)`
Vào bài toán ta có:
`sqrt{1+1/1^2+1/2^2}+sqrt{1+1/2^2+1/3^2}+....+sqrt{1+1/11^2+1/12^2}`
`=1+1-1/2+1+1/2-1/3+...+1+1/11-1/12`
`=11.1-1/12`
`=11-1/12=131/12`
4) Ta có: \(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+...+\sqrt{1+\dfrac{1}{11^2}+\dfrac{1}{12^2}}\)
\(=1+\dfrac{1}{1}-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{11}-\dfrac{1}{12}\)
\(=12-\dfrac{1}{12}=\dfrac{143}{12}\)
