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72^45-72^44=72^44(72-1)=72^44*71
72^44-72^43=72^43(72-1)=72^43*71
=>72^45-72^44>72^44-72^43
12.x-27:24 = 28
12.x-\(\frac{9}{8}\)= 28
12x = 28+\(\frac{9}{8}\)
12x = \(29\frac{1}{8}\)
x = \(29\frac{1}{8}:12\)
x = \(\frac{233}{96}\)
\(a.10^{30}=\left(10^3\right)^{10}=1000^{10}\\ 2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì 100010 < 102410 => 1030 < 2100
\(b,333^{444}=\left(111\cdot3\right)^{444}=111^{444}\cdot3^{444}=111^{444}\cdot81^{111}\\ 444^{333}=\left(111\cdot4\right)^{333}=111^{333}\cdot4^{333}=111^{333}\cdot64^{111}\)
Vì 111444 >111333 ; 81111 > 64111 => 333444 > 444333
\(B=1996.2000=\left(1998-2\right)\left(1998+2\right)=1998^2+2.1998-2.1998-2^2=1998^2-4< 1998^2=1998.1998=A\)
\(B=1996\cdot2000\)
\(=\left(1998-2\right)\left(1998+2\right)\)
\(=1998^2-4< 1998^2\)
hay A>B
Vậy: A lớn hơn B 4 đơn vị
A = 1-1/2011+1-1/2012 = 2-(1/2011+1/2012) > 1 ( vì 1/2011+1/2012 < 1 )
B = 4021/4023 = 1-2/4023 < 1
=> A > B
k mk nha
\(2^{10}=1024< 1029=3.7^3\)
\(\Leftrightarrow\left(2^{10}\right)^{238}< \left(3.7^3\right)^{238}\)
\(\Leftrightarrow2^{2380}< 3^{238}.7^{714}\) \(\left(1\right)\)
\(3^5=243< 256=2^8\) \(\left(2\right)\)
\(3^3=27< 32=2^5\) \(\left(3\right)\)
Từ \(\left(2\right)\), \(\left(3\right)\) ta có:
\(3^{328}=3^3.3^{325}=3^3\left(3^5\right)^{47}< 2^5\left(2^8\right)^{47}=2^{381}\)\(\left(4\right)\)
Từ \(\left(1\right)\), \(\left(4\right)\) ta có:
\(2^{2380}< 3^{238}.7^{714}\)
\(\Leftrightarrow2^{2380}< 2^{381}.7^{714}\)
\(\Leftrightarrow2^{1999}< 7^{714}\)
\(\Leftrightarrow2^{1993}< 7^{714}\).