tính giá trị của biểu thức sau:
A=\(\frac{2^{10}.2^{13}-2^{21}.20}{2^{24}}\)
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Tính giá trị biểu thức sau bằng cách hợp lí:
A= 1/2 + (-1/7) - (-1/13) +-1/13 - (-2/5) + -11/21 +1/10
\(A=\frac{1}{2}+\frac{-1}{7}-\frac{-1}{13}+\frac{-1}{13}-\frac{-2}{5}+\frac{-11}{21}+\frac{1}{10}\)
\(A=\frac{5}{10}-\frac{3}{21}+\frac{1}{13}-\frac{1}{13}+\frac{4}{10}-\frac{11}{21}+\frac{1}{10}\)
\(A=\left(\frac{5+4+1}{10}\right)+\left(\frac{-3}{21}-\frac{11}{21}\right)+\left(\frac{1}{13}-\frac{1}{13}\right)\)
\(A=1+\frac{-2}{3}=\frac{3-2}{3}=\frac{1}{3}\)
\(A=\frac{10-1\frac{1}{6}\times\frac{6}{7}}{21:\frac{11}{2}+5\frac{2}{11}}\)
\(A=\frac{10-\frac{7}{6}\times\frac{6}{7}}{21:\frac{11}{2}+\frac{57}{11}}\)
\(A=\frac{10-1}{\frac{42}{11}+\frac{57}{11}}\)
\(A=\frac{9}{9}=1\)
a/\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
= \(\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{1-3}{1+5}=\frac{-2}{6}=-3\)
\(a,\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)
\(=>\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{3}{7}\right)-\frac{20}{15}\)
\(=>1+\frac{16}{21}-\frac{20}{15}\)
\(=>\frac{37}{21}-\frac{20}{15}\)
\(=>\frac{3}{7}\)
\(b,12-8\cdot\left(\frac{3}{2}\right)^3\)
\(=>12-8\cdot\frac{27}{8}\)
\(=>12-27\)
\(=>-15\)
\(c,\left(\frac{1}{9}\right)^{2005}\cdot9^{2005}-96^2:24^2\)
\(=>\left(\frac{1^{2005}^{ }}{9^{2005}}\cdot9^{2005}\right)-\left(96^2:24^2\right)\)
\(=>\left(1^{2005}\right)-16\)
\(=>1-16\)
\(=>-15\)
\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}=3\)
\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}=\frac{2^2.78}{104}=\frac{2^2.2.39}{2^3.13}=\frac{2^3.39}{2^3.13}=3\)
B3:\(\Rightarrow90.10^n-10^n.10^2+10^n.10-20\Rightarrow10^n.\left(90-10^2\right)+10^n.10-20\)
\(\Rightarrow10^n.\left(90-100\right)+10^n.10-20\Rightarrow-10.10^n+10^n.10-20\Rightarrow-20\)
\(A=-\left(x^2-x+5\right)=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{19}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\right]\)
\(=-\left(x-\frac{1}{2}\right)^2-\frac{19}{4}\le-\frac{19}{4}\)
Vậy \(A_{min}=-\frac{19}{4}\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
\(A=\frac{2^{23}-2^{21}\cdot2^2\cdot5}{2^{24}}\)
\(A=\frac{2^{23}-2^{23}\cdot5}{2^{24}}\)
\(A=\frac{2^{23}\left(1-5\right)}{2^{24}}\)
\(A=\frac{-\left(2^{23}\cdot2^2\right)}{2^{24}}=-\frac{2^{25}}{2^{24}}=-2\)
cảm ơn bạn