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b: \(=32+\dfrac{5}{16}-3\cdot32\)
=-64+5/16
=-1019/16
27 : 22 = 25 = 32
54 : 53 = 5
24 - 3.25 = 16 - 96 = - 80
\(a,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256,2^9=512,2^{10}=1024\)
\(b,3^2=9,3^3=27,3^4=81,3^5=243\)
\(c,4^2=16,4^3=64,4^4=256\)
\(d,5^2=25,5^3=125,5^4=625\)
a: \(2^3-5^3:5^2+12\cdot2^2\)
\(=8-5+48\)
\(=51\)
b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5\)
\(=5\cdot\left[10+90\right]-5\)
=495
\(A=\frac{54.107-53}{53.107+54}=\frac{54.107+54-107}{\left(54-1\right).107+54}=\frac{54.\left(107+1\right)-107}{54.107-107+54}=\frac{54.108-107}{54.\left(107+1\right)-107}=\frac{54.108-107}{54.108-107}=1\)
\(B=\frac{135.296-133}{134.269+135}=\frac{135.296+135-268}{\left(135-1\right).269+135}=\frac{135.\left(296+1\right)-268}{135.269-269+135}=\frac{135.297-268}{135.\left(269+1\right)-269}=\frac{135.297-268}{135.270-269}>1\)
\(\Rightarrow A< B\)
\(A=\frac{54.107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.296-133}{134.269+135}=\frac{134.296+296-133}{134.269+135}=\frac{134.296+163}{134.269+135}\)
Vì \(134=134;296>269\Rightarrow134.269< 134.296\)mà\(163>135\Rightarrow134.269+135< 134.296+163\)hay \(B>1\)
Ta có \(A=1;B>1\Rightarrow A< B\)
\(2^7:2^2+5^4:5^3:2^4-3.2^5\)
\(=2^5+\frac{5}{16}-96\)
\(=\frac{-1019}{16}\)