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ta có : 4^40 = ( 4^8)^5 = 65536^5

5^35 = ( 5^7 )^5 = 78125^5

Vì 65536^5 < 78125^5 => 4^40 < 

tk cho mk mk làm đầy đủ cho 

viết tiếp : 4^40 < 5^35

Ta có: \(\left(\frac{-1}{4}\right)^{40}=\left[\left(\frac{-1}{4}\right)^2\right]^{20}=\left(\frac{1}{16}\right)^{20}\)

             \(\left(\frac{-1}{5}\right)^{34}=\left[\left(\frac{-1}{5}\right)^2\right]^{17}=\left(\frac{1}{25}\right)^{17}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{20}>\left(\frac{1}{25}\right)^{17}\)

Vậy \(\left(\frac{-1}{4}\right)^{40}>\left(\frac{-1}{5}\right)^{34}\)

\(\left(\frac{-1}{40}\right)^{40}< \left(\frac{-1}{5}\right)^{34}\)

\(\left(-\frac{1}{4}\right)40>\left(-\frac{1}{5}\right)34\)

chuc bn hoc gioi!

nhaE$

Tobot Z hihi

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mình thi rồi.lớp 7 vòng 5

Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)

Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)

\(\dfrac{1}{32}>\dfrac{1}{40}\)

\(\dfrac{1}{33}>\dfrac{1}{40}\)

\(\dfrac{1}{34}>\dfrac{1}{40}\)

\(\dfrac{1}{35}>\dfrac{1}{40}\)

\(\dfrac{1}{36}>\dfrac{1}{40}\)

\(\dfrac{1}{37}>\dfrac{1}{40}\)

\(\dfrac{1}{38}>\dfrac{1}{40}\)

\(\dfrac{1}{39}>\dfrac{1}{40}\)

\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)

Vậy \(S>\dfrac{1}{4}\)

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