Question

so sánh P và Q biết P = \(6^{2016}+4\) /\(6^{2016}-1\) và Q = \(6^{2016}\) / \(6^{2016}-5\)

Answer 1

Ta có :

\(P=\dfrac{6^{2016}+4}{6^{2016}-1}=\dfrac{6^{2016}-1+5}{6^{2016}-1}=\dfrac{6^{2016}-1}{6^{2016}-1}+\dfrac{5}{6^{2016}-1}\)\(=1+\dfrac{5}{6^{2016}-1}\)

\(Q=\dfrac{6^{2016}}{6^{2016}-5}=\dfrac{6^{2016}-5+5}{6^{2016}-5}=\dfrac{6^{2015}-5}{6^{2016}-5}+\dfrac{5}{6^{2016}-5}=1+\dfrac{5}{6^{2016}-5}\)

Vì \(1+\dfrac{5}{6^{2016}-1}< 1+\dfrac{5}{6^{2016}-5}\Rightarrow P< Q\)

Answer 2

Ta có:

\(P-Q=\dfrac{6^{2016}+4}{6^{2016}-1}-\dfrac{6^{2016}}{6^{2016}-5}=1+\dfrac{5}{6^{2016}-1}-1-\dfrac{5}{6^{2016}-5}\)

\(=\dfrac{5}{6^{2016}-1}-\dfrac{5}{6^{2016}-5}=5\left(\dfrac{1}{6^{2016}-1}-\dfrac{1}{6^{2016}-5}\right)< 0\)

Vậy A < B

Answer 3

Ta có \(P=\dfrac{6^{2016}+4}{6^{2016}-1}>\dfrac{6^{2016}+4-4}{6^{2016}-1-4}=\dfrac{6^{2016}}{6^{2016}-5}=Q\)

Vậy P<Q

Có cần công thức chứng minh \(\dfrac{a}{b}>\dfrac{a-m}{b-m}\)không vậy.Mình biết.