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Hãy luôn nhớ cảm ơn và vote 5*
nếu câu trả lời hữu ích nhé!
`P=\sqrt{1+1/2^2+1/3^2}+\sqrt{1+1/3^2+1/4^2}+...+\sqrt{1+1/(n+2)^2+1/(n+3)^2}`
`P=\sqrt{(2^2 .3^2+3^2+2^2)/(2^2 .3^2)}+\sqrt{(3^2 .4^2+4^2+3^2)/(3^2 .4^2)}+...+\sqrt{((n+2)^2 .(n+3)^2+(n+3)^2+(n+2)^2)/((n+2)^2 .(n+3)^2)}`
Ta có:
`(n+2)^2 .(n+3)^2+(n+3)^2+(n+2)^2`
`=(n+2)^2 .[(n+2)+1]^2+(n+3)^2+(n+2)^2`
`=(n+2)^2 .[(n+2)^2+2(n+2)+1]+(n+3)^2+(n+2)^2`
`=(n+2)^4+2(n+2)^3+(n+2)^2+(n+3)^2+(n+2)^2`
`=(n+2)^4+[2(n+2)^3+2(n+2)^2]+(n+3)^2`
`=(n+2)^4+2(n+2)^2(n+2+1)+(n+3)^2`
`=(n+2)^4+2(n+2)^2(n+3)+(n+3)^2`
`=[(n+2)^2+n+3]^2`
`->P=\sqrt{(2^2+3)^2/(2^2 .3^2)}+\sqrt{(3^2+4)^2/(3^2 .4^2)}+...+\sqrt{[(n+2)^2+n+3]^2/((n+2)^2 .(n+3)^2)}`
`P=(2^2+3)/(2.3)+(3^2+4)/(3.4)+...+((n+2)^2+n+3)/((n+2)(n+3))`
`P=2^2/(2.3)+3/(2.3)+3^2/(3.4)+4/(3.4)+...+(n+2)^2/((n+2)(n+3))+(n+3)/((n+2)(n+3))`
`P=2/3+1/2+3/4+1/3+...+(n+2)/(n+3)+1/(n+2)`
`P=1/2+(2/3+1/3)+(3/4+1/4)+...+((n+2)/(n+3)+1/(n+3))+(1/(n+2)+(n+1)/(n+2))`
`P=1/2+3/3+4/4+...+(n+3)/(n+3)+(n+2)/(n+2)`
`P=1/2+\underbrace{1+1+...+1+1}_{\text{n chữ số 1}}`
`P=1/2+n`
Vậy `P=1/2+n`
Hãy giúp mọi người biết câu trả lời này thế nào?
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