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\(\left(\frac{1}{3}\right)^{500}=\left(\frac{1}{3}^5\right)^{100}=\frac{1}{243}^{100}\)
\(\left(\frac{1}{5}\right)^{300}=\left(\frac{1}{5}^3\right)^{100}=\frac{1}{125}^{100}\)
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 \(2^{90}>5^{36}\)
\(A=x+\left(x+\frac{1}{5}\right)+\left(x+\frac{2}{5}\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{4}{5}\right)\)
\(=5x+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\)
\(=5x+2\)
\(B=5x\)
\(\Rightarrow A>B\)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!
Ta có:
(-1/5)300 = (-1)300/5300 = 1/(53)100 = 1/125100
(-1/3)500 = (-1)500/3500 = 1/(35)100 = 1/243100
Vì 125100 < 243100
=> 1/125100 > 1/243100
=> (-1/5)300 > (-1/3)500
Ta có : \(\left(-\frac{1}{5}\right)^{300}=\left(-\frac{1}{5}\right)^{3.100}=\left(-\frac{1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)
\(\left(-\frac{1}{3}\right)^{500}=\left(-\frac{1}{3}\right)^{5.100}=\left(-\frac{1}{243}\right)^{100}=\left(\frac{1}{243}\right)^{100}\)
Mà \(125< 243\Rightarrow\frac{1}{125}>\frac{1}{243}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{243}\right)^{100}\)
\(=>\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)