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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 )
\(B=4\cdot\left(-\frac{1}{2}\right)^3:\left(\frac{4}{5}\right)^0\cdot\frac{1}{2}-\frac{\frac{3}{5}-\frac{3}{9}+\frac{3}{13}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{13}}\)
\(=4\cdot\frac{-1}{8}:1\cdot\frac{1}{2}-\frac{3\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{13}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{13}\right)}\)
\(=-\frac{1}{4}-\frac{3}{7}=-\frac{19}{28}\)
\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)
\(=\left(\frac{3}{7}\right)^9\)
\(\left(\frac{1}{6}\right)^5+\left(\frac{1}{8}\right)^3=\frac{1}{2^5}.\frac{1}{3^5}+\frac{1}{2^9}=\frac{1}{2^5}\left(\frac{1}{3^5}+\frac{1}{2^4}\right)=\frac{259}{124416}\)
Chúc bạn
học tốt!!!!!!!!!!!!!!!
\(\left(\frac{1}{6}\right)^5+\left(\frac{1}{8}\right)^3\)
= \(\frac{1}{7776}+\frac{1}{512}\)
= \(\frac{16}{124416}+\frac{243}{124416}\)
= \(\frac{259}{124416}\)
a) Lm r nkoa!
b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)
\(=\left(0,25x\right):3=\frac{20}{3}\)
\(\Rightarrow0,25x=\frac{20}{3}\cdot3=20\)
\(\Rightarrow x=20:0,25=80\)
\(\Rightarrow x=80\)
c) \(0,01:2,5=\left(0,75x\right):0,75\)
\(=\frac{1}{250}=\left(0,75x\right):0,75\)
\(\Rightarrow0,75x=\frac{1}{250}\cdot0,75=\frac{3}{1000}\)
\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)
\(\Rightarrow x=\frac{1}{250}\)
d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(=\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow0,1x=\frac{2}{3}:\frac{5}{3}=\frac{2}{5}\)
\(\Rightarrow x=\frac{2}{5}:0,1=4\)
\(\Rightarrow4\)
a, => |5/3.x| = 1/6
=> 5/3.x = -1/6 hoặc 5/3.x = 1/6
=> x = -1/10 hoặc x = 1/10
Tk mk nha
\(\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}=\frac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}}=\frac{40.\left(-15\right).64}{135.\left(-2\right)^{14}}=\frac{5.8.3.\left(-5\right).64}{15.9.\left(-2\right)^{14}}=\frac{8.\left(-5\right).\left(-2\right)^6}{9.\left(-2\right)^{14}}=\frac{\left(-2\right)^3.5}{9.\left(-2\right)^8}=\frac{5}{9.\left(-2\right)^5}\)
\(\dfrac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}\) =\(\dfrac{18.34-18.124}{9.\left(-4\right).17+9.\left(-52\right)}=\dfrac{18.\left(34-124\right)}{9.\left(-68\right)+9.\left(-52\right)}\) =\(\dfrac{18.\left(-90\right)}{9.\left(-68-52\right)}=\dfrac{18.\left(-90\right)}{9.\left(-120\right)}=\dfrac{3}{2}\)
\(\frac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}=\frac{18.34-18.124}{9.\left(-4\right)+17+9.\left(-52\right)}=\frac{18.\left(34-124\right)}{9.\left(17-52\right).\left(-4\right)}\)
\(\frac{18.\left(-90\right)}{9.\left(-35\right).\left(-4\right)}=\frac{18.\left(-90\right)}{9.140}=\frac{9.2.\left(-9\right).10}{9.7.2.10}=\frac{-9}{7}\)