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Chắc đấy! Tick nhé!

\(\left(\frac{1}{3}\right)^{500}=\left(\frac{1}{3}^5\right)^{100}=\frac{1}{243}^{100}\)

\(\left(\frac{1}{5}\right)^{300}=\left(\frac{1}{5}^3\right)^{100}=\frac{1}{125}^{100}\)

Vì \(\frac{1}{243}<\frac{1}{125}=>\frac{1}{243}^{100}<\frac{1}{125}^{100}=>\left(\frac{1}{3}\right)^{500}<\left(\frac{1}{5}\right)^{300}\)

3^-500=(3⁵)^-100= 243^-100

5^-300= (5³)^-100 =125^-100

^{243>125 =>}    243^-100<125^-100

Hay 3^-500 <5^-300

a) Ta có :\(\left(\frac{-1}{5}\right)^{300}=\frac{-1^{300}}{5^{300}}=\frac{1}{125^{100}}\)

\(\left(\frac{-1}{3}\right)^{500}=\frac{-1^{500}}{3^{500}}=\frac{1}{243^{100}}\)

Mà \(\frac{1}{125^{100}}>\frac{1}{243^{100}}\)

\(\Rightarrow\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)

b)Ta có :\(2^{90}=\left(2^{15}\right)^6=32768^6\)

\(5^{36}=\left(5^6\right)^6=15625^6\)

Vì \(32768^6>15625^6\Rightarrow2^{90}>5^{36}\)

a.Ta có: \(\left(\frac{-1}{5}\right)^{300}=\left(\frac{-1}{5}^3\right)^{100}=\left(\frac{-1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)

\(\left(\frac{-1}{3}\right)^{500}=\left(\frac{-1}{3}^5\right)^{100}=\left(\frac{-1}{243}\right)^{100}=\left(\frac{1}{234}\right)^{100}\)

Mà: \(\frac{1}{125}>\frac{1}{234}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{234}\right)^{100}\)

Vậy \(\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)

b.Ta có: \(2^{90}=\left(2^{10}\right)^9=1024^9\)

\(5^{36}=\left(5^4\right)^9=625^9\)

Mặt khác: \(1024>625\Rightarrow1024^9>625^9\)

 Vậy \(2^{90}>5^{36}\)

x<y nhé bạn :)

Ta có : (-1/5)^300=(-1/5^3)100=(-1/125)^100

(-1/3)^500=(-1/3^5)^100=(-1/243)^100

vì (-1/243)^100<(-1/125)^100→(-1/5)^300>(-1/3)^500

b, ta có:-(-2)^300=(2^3)^100=8^100

(-3)^200=(-3^2)^100=9^100

vì 8^100<9^100→-(-2)^300<(-3)^200

\(A=x+\left(x+\frac{1}{5}\right)+\left(x+\frac{2}{5}\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{4}{5}\right)\)

\(=5x+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\)

\(=5x+2\)

\(B=5x\)

\(\Rightarrow A>B\)Với \(\forall\)\(x\)

#)Giải :

\(A=\left[x\right]+\left[1+\frac{1}{5}\right]+\left[x+\frac{2}{5}\right]+\left[x+\frac{3}{5}\right]+\left[x+\frac{4}{5}\right]\)

Thay x = 3,7 vào biểu thức, ta có :

\(A=\left[3,7\right]+\left[3,7+\frac{1}{5}\right]+\left[3,7+\frac{2}{5}\right]+\left[3,7+\frac{3}{5}\right]+\left[3,7+\frac{4}{5}\right]\)

\(A=\left[3,7+3,7+3,7+3,7+3,7\right]+\left[1+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right]\)

\(A=18,5+3\)

\(A=21,5\)

\(B=\left[5x\right]=\left[5\times3,7\right]=18,5\)

Vì 21,5 > 18,5 \(\Rightarrow A>B\)

Ta có:

\(\left(-\frac{1}{5}\right)^{300}=\left[\left(-\frac{1}{5}\right)^3\right]^{100}=\left(-\frac{1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}.\)

\(\left(-\frac{1}{3}\right)^{500}=\left[\left(-\frac{1}{3}\right)^5\right]^{100}=\left(-\frac{1}{243}\right)^{100}=\left(\frac{1}{243}\right)^{100}.\)

Vì \(125< 243\) nên \(\frac{1}{125}>\frac{1}{243}.\)

\(\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{243}\right)^{100}\)

\(\Rightarrow\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}.\)

Chúc bạn học tốt!

thanks nhiều