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a: Ta có: \(\dfrac{1}{2}=0.5\)
\(\dfrac{3}{4}=0.75\)
mà 0,5<0,75
nên x<y
a) \(\frac{-2}{3}>\frac{4}{-5}\)
b) \(\frac{-18}{31}< \frac{-18.18.18}{31.31.31}\)
c) \(\frac{-13}{31}< \frac{29}{-88}\)
\(\left(\frac{1}{2}\right)^{40}=\left(\frac{1}{2}\right)^{10\cdot4}=\left(\frac{1}{16}\right)^{10}\)
Mà ta có
\(\left(\frac{1}{32}\right)^{10}< \left(\frac{1}{16}\right)^{10}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{40}>\left(\frac{1}{32}\right)^{10}\)
𝓣𝓪 𝓬𝓸́: \(1-\dfrac{2002}{2003}=\dfrac{1}{2003}\)
\(1-\dfrac{2003}{2004}=\dfrac{1}{2004}\)
𝓓𝓸 \(\dfrac{1}{2003}>\dfrac{1}{2004}\)
𝓷𝓮̂𝓷 \(\dfrac{2002}{2003}>\dfrac{2003}{2004}\)
𝓥𝓪̣̂𝔂 \(\dfrac{2002}{2003}>\dfrac{2003}{2004}\)
Ta có :
\(\dfrac{2002}{2003}< \dfrac{2002+1}{2003+1}=\dfrac{2003}{2004}\)
Vậy \(\dfrac{2002}{2003}< \dfrac{2003}{2004}\)
Bài 1:
ta có: 333<3333; 444<4444
=> 333444<33334444
Bài 2:
\(A=\frac{21^5}{81}=\frac{\left(3.7\right)^5}{3^4}=\frac{3^5.7^5}{3^4}=3.7^5=50421\)
\(B=\frac{3^3.\left(0,5\right)^5}{\left(1,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{\left(3.0,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{3^5.\left(0,5\right)^5}=\frac{1}{3^2}=\frac{1}{9}\)
\(C=2^2.\frac{1}{128}.45.2^{-6}=\frac{2^2.45}{128.64}=\frac{2^2.45}{2^7.2^6}=\frac{45}{2^{11}}=\frac{45}{2048}\)
\(D=\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+2^2.3^3+3^3}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}\)\(=3^3.\left(-1\right)=-27\)
Bài 1:
a: Sửa đề: 1/3^200
1/2^300=(1/8)^100
1/3^200=(1/9)^100
mà 1/8>1/9
nên 1/2^300>1/3^200
b: 1/5^199>1/5^200=1/25^100
1/3^300=1/27^100
mà 25^100<27^100
nên 1/5^199>1/3^300
\(\dfrac{105}{-15}\) = -7 > - 7,112
Vậy \(\dfrac{105}{-15}\) > -7,112
105/-15 va -7,112
Ta co : 105/-15 = -7
Ma -7 < -7,112
Vay 105/-15 <-7,112
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-1/3=0,(3)>-0,333