Tìm tỉ số của A và B biết:
A=\(\frac{\left(-2\right)^0+1^{2017}+\left(\frac{-1}{3}\right)^8.3^8}{2^{15}}\)
B=\(\frac{6^2}{2^{16}}\)
Help me please! Thanks !
K
Khách
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24 tháng 10 2016
\(A=\frac{\left(-2\right)^0+1^{2017}+\left(-\frac{1}{3}\right)^8.3^8}{2^{15}}\)
\(=\frac{1+1+\frac{1}{3^8}.3^8}{2^{15}}\)
\(=\frac{1+1+1}{2^{15}}\)
\(=\frac{3}{2^{15}}\)
\(B=\frac{6^2}{2^{16}}\)
\(=\frac{2^2.3^2}{2^2.2^{14}}\)
\(=\frac{9}{2^{14}}\)
Dễ dàng thấy \(9>3\)
\(2^{14}< 2^{15}\)
Phép chia có cùng mẫu, tử lớn hơn thì đã lớn hơn, nay mẫu còn nhỏ hơn, chắc chắn rằng \(B>A\)
Vậy ...
T
27 tháng 5 2020
Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)
\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)
Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Ai có cách hay?
T
27 tháng 5 2020
1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.
2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)
\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)
\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)
23 tháng 4 2019
a,\(\frac{4}{9}.\frac{2}{6}=\frac{4}{27}\)
b,\(1\frac{1}{3}.\left(0,5\right)+\left(\frac{8}{15}-\frac{19}{30}\right):\frac{6}{15}\)
=\(\frac{4}{3}.\frac{1}{2}+\left(\frac{16}{30}-\frac{19}{30}\right).\frac{15}{6}\)
=\(\frac{2}{3}+\frac{-1}{10}.\frac{15}{6}\)
=\(\frac{2}{3}+\frac{-1}{4}\)
=\(\frac{8}{12}+\frac{-3}{12}=\frac{5}{12}\)
bài2
a,\(\left(\frac{2}{7}.x+\frac{3}{7}\right):2\frac{1}{5}-\frac{3}{7}=1\)
=>\(\left(\frac{2}{7}.x+\frac{3}{7}\right):\frac{11}{5}=1+\frac{3}{7}=\frac{10}{7}\)
=>\(\frac{2}{7}.x+\frac{3}{7}=\frac{10}{7}.\frac{11}{5}\)
=>\(\frac{2}{7}.x+\frac{3}{7}=\frac{22}{7}\)
=>\(\frac{2}{7}.x=\frac{22}{7}-\frac{3}{7}=\frac{19}{7}\)
=>\(x=\frac{19}{7}:\frac{2}{7}=\frac{19}{7}.\frac{7}{2}=\frac{19}{2}\)
vậy x\(=\frac{19}{2}\)
17 tháng 7 2016
Cũng khuya rồi , mình làm câu 1 thôi nhé !
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{2.5^{22}-9.5^{21}}{\left(5^2\right)^{10}}\)
\(\frac{5^{21}.\left(2.5-9\right)}{5^{20}}=5.\left(10-9\right)=5\)
15 tháng 7 2019
\(a,\left[\left(-\frac{1}{2}\right)^3-\left(\frac{3}{4}\right)^3.\left(-2\right)^2\right]:\left[2.\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}\right]\)
\(=\left[\left(-\frac{1}{8}\right)-\frac{27}{64}.4\right]:\left[2.\left(-1\right)+\frac{9}{16}-\frac{3}{8}\right]\)
\(=\left[\left(-\frac{1}{8}-\frac{27}{16}\right)\right]:\left[-2+\frac{9}{16}-\frac{3}{8}\right]\)
\(=\frac{-2-27}{16}:\frac{-32+9-6}{16}\)
\(=-\frac{29}{16}:\frac{-29}{16}=1\)
\(b,\left[\left(\frac{4}{3}\right)^{-2}\left(\frac{3}{2}\right)^4\right]:\left(\frac{3}{2}\right)^6\)
\(=\left(\frac{9}{16}.\frac{81}{16}\right):\frac{729}{64}\)
\(=\frac{729}{64}:\frac{729}{64}=1\)
Ta có :
\(\frac{A}{B}=\frac{\left(-2\right)^0+1^{2017}+\left(\frac{-1}{3}\right)^8.3^8}{2^{15}}:\frac{6^2}{2^{16}}\)
=> \(\frac{A}{B}=\frac{1+1+\left(\frac{-1}{3}.3\right)^8}{2^{15}}.\frac{2^{16}}{6^2}\)
=> \(\frac{A}{B}=\frac{1+1+1^8}{1}.\frac{2}{6^2}\)
=> \(\frac{A}{B}=\frac{3}{1}.\frac{2}{2^2.3^2}\)
=> \(\frac{A}{B}=\frac{1}{2.3}=\frac{1}{6}\)
Ta có:
\(\frac{A}{B}\)=\(\frac{\left(-2\right)^0+1^{2017}+\left(\frac{-1}{3}\right)^8\cdot3^8}{2^{15}}\):\(\frac{6^2}{2^{16}}\)
=>\(\frac{A}{B}\)=\(\frac{1+1+\left(\frac{-1}{3}\cdot3\right)^8}{2^{15}}\).\(\frac{2^{16}}{6^2}\)
=>\(\frac{A}{B}\)=\(\frac{1+1+1^8}{2^{15}}\).\(\frac{2^{16}}{6^2}\)
=>\(\frac{A}{B}\)=\(\frac{3}{2^{15}}\).\(\frac{2^{16}}{6^2}\)
=>\(\frac{A}{B}\)=\(\frac{2}{3.2^2}\)
=>\(\frac{A}{B}\)=\(\frac{1}{6}\)