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\(B=4\cdot\left(-\frac{1}{2}\right)^3:\left(\frac{4}{5}\right)^0\cdot\frac{1}{2}-\frac{\frac{3}{5}-\frac{3}{9}+\frac{3}{13}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{13}}\)
\(=4\cdot\frac{-1}{8}:1\cdot\frac{1}{2}-\frac{3\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{13}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{13}\right)}\)
\(=-\frac{1}{4}-\frac{3}{7}=-\frac{19}{28}\)
c) C = ( 1 - 2 ) + ( 3 - 4 ) + ... + ( 79 - 80 )
C = ( -1 ) + ( -1 ) + ... + ( -1 )
C = ( -1 ) x ( 80 - 1 + 1 ) : 2
C = ( -1 ) x 80 : 2
C = ( -40 )
\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)
\(=\left(\frac{3}{7}\right)^9\)
\(\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}=\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}}=\frac{40.\left(-15\right).64}{135.\left(-2\right)^{14}}=\frac{5.8.3.\left(-5\right).64}{15.9.\left(-2\right)^{14}}=\frac{8.\left(-5\right).\left(-2\right)^6}{9.\left(-2\right)^{14}}=\frac{\left(-2\right)^3.5}{9.\left(-2\right)^8}=\frac{5}{9.\left(-2\right)^5}\)
\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)
\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)
mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)
\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)
Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)
\(\Rightarrow A>-\dfrac{11}{21}\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)
\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)
Dễ thấy A có 9 thừa số, suy ra
\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)
Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)
Vậy \(A< \dfrac{-11}{21}\)
\(\left(\frac{1}{6}\right)^5+\left(\frac{1}{8}\right)^3=\frac{1}{2^5}.\frac{1}{3^5}+\frac{1}{2^9}=\frac{1}{2^5}\left(\frac{1}{3^5}+\frac{1}{2^4}\right)=\frac{259}{124416}\)
Chúc bạn
học tốt!!!!!!!!!!!!!!!
\(\left(\frac{1}{6}\right)^5+\left(\frac{1}{8}\right)^3\)
= \(\frac{1}{7776}+\frac{1}{512}\)
= \(\frac{16}{124416}+\frac{243}{124416}\)
= \(\frac{259}{124416}\)
\(\dfrac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}\) =\(\dfrac{18.34-18.124}{9.\left(-4\right).17+9.\left(-52\right)}=\dfrac{18.\left(34-124\right)}{9.\left(-68\right)+9.\left(-52\right)}\) =\(\dfrac{18.\left(-90\right)}{9.\left(-68-52\right)}=\dfrac{18.\left(-90\right)}{9.\left(-120\right)}=\dfrac{3}{2}\)
\(\frac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}=\frac{18.34-18.124}{9.\left(-4\right)+17+9.\left(-52\right)}=\frac{18.\left(34-124\right)}{9.\left(17-52\right).\left(-4\right)}\)
\(\frac{18.\left(-90\right)}{9.\left(-35\right).\left(-4\right)}=\frac{18.\left(-90\right)}{9.140}=\frac{9.2.\left(-9\right).10}{9.7.2.10}=\frac{-9}{7}\)