Question

[d b t]

7¹⁰​+1/7¹⁰ - 1 and 7¹⁰-1/7¹⁰ -3

đề là so sánh

Answer 1

[Đinh Đức Hùng]

Ta có :

\(\frac{7^{10}+1}{7^{10}-1}=\frac{\left(7^{10}-1\right)+2}{7^{10}-1}=1+\frac{2}{7^{10}-1}\)

\(\frac{7^{10}-1}{7^{10}-3}=\frac{\left(7^{10}-3\right)+2}{7^{10}-3}=1+\frac{2}{7^{10}-3}\)

Vì \(7^{10}-1>7^{10}-3\) nên \(\frac{2}{7^{10}-1}< \frac{2}{7^{10}-3}\) \(\Rightarrow1+\frac{2}{7^{10}-1}< 1+\frac{2}{7^{10}-3}\)

Do đó :\(\frac{7^{10}+1}{7^{10}-1}< \frac{7^{10}-1}{7^{10}-3}\)

Answer 2

[Đỗ Việt Nhật]

\(\frac{7^{10}+1}{7^{10}-1}\)

=1+\(\frac{2}{7^{10}-1}\)

\(\frac{7^{10}-1}{7^{10}-3}\)

=1+\(\frac{2}{7^{10}-3}\)

vì \(7^{10}\)-1>\(7^{10}\)-3

nên \(\frac{2}{7^{10}-1}\)<\(\frac{2}{7^{10}-3}\)

\(\Rightarrow\)1+\(\frac{2}{7^{10}-1}\)<1+\(\frac{2}{7^{10}-3}\)

Vậy\(\frac{7^{10}+1}{7^{10}-1}\)<\(\frac{7^{10}-1}{7^{10}-3}\)