Chứng minh rằng : $\sqrt[3]{n}$ > $\sqrt[4]{n+1}$ trong đó n thuộc N ,n$\geq$ 3 07/07/2021 Bởi Iris Chứng minh rằng : $\sqrt[3]{n}$ > $\sqrt[4]{n+1}$ trong đó n thuộc N ,n$\geq$ 3
Đáp án: `\root{3}{n}>\root{4}{n+1}` `<=>(\root{3}{n})^{12}>(\root{4}{n+1})^{12}` `<=>n^4>(n+1)^3` `<=>n^4>n^3+3n^2+3n+1` Ta có:`n^4=n.n^3>=3n^3` `3n^3=n^3+2n^3=n^3+n.2n^2>=n^3+6n^2` `=>n^4>=n^3+6n^2` `<=>n^4>=n^3+3n^2+n.3n>=n^3+3n^2+9n` `<=>n^4>=n^3+3n^2+3n+6n>n^3+3n^2+3n+1` `=>n^4>n^3+3n^2+3n+1` Hay `\root{3}{n}>\root{4}{n+1}AAn in NN,n>=3` Bình luận
Đáp án:
`\root{3}{n}>\root{4}{n+1}`
`<=>(\root{3}{n})^{12}>(\root{4}{n+1})^{12}`
`<=>n^4>(n+1)^3`
`<=>n^4>n^3+3n^2+3n+1`
Ta có:`n^4=n.n^3>=3n^3`
`3n^3=n^3+2n^3=n^3+n.2n^2>=n^3+6n^2`
`=>n^4>=n^3+6n^2`
`<=>n^4>=n^3+3n^2+n.3n>=n^3+3n^2+9n`
`<=>n^4>=n^3+3n^2+3n+6n>n^3+3n^2+3n+1`
`=>n^4>n^3+3n^2+3n+1`
Hay `\root{3}{n}>\root{4}{n+1}AAn in NN,n>=3`