Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)(-7 / 5 + 3 / 8 ) : 2009 / 2010 + (-3 / 5+5 / 8) : 2009 / 2010
=[ (-7 / 5 + 3 / 8 ) + (-3 / 5+5 / 8) ] : 2009/2010
=[ -7/5 +3/8 + (-3)/5+5/8 ] : 2009/2010
=[ (-7/5 + (-3)/5) + (3/8 + 5/8) ] :2009/2010
=[-2+1] : 2009/2010
=-1 :2009/2010
=-2009/2010
b)\(\frac{9^8\cdot4^3}{27^4\cdot6^5}=\frac{\left(3^2\right)^8\cdot\left(2^2\right)^3}{\left(3^3\right)^4\cdot\left(2\cdot3\right)^5}=\frac{3^{16}\cdot2^6}{3^{12}\cdot2^5\cdot3^5}=\frac{3^{16}\cdot2^5\cdot2}{3^{16}\cdot3^1\cdot2^5}=\frac{2}{3^1}=\frac{2}{3}\)
\(\left(5-\frac{2}{3}+\frac{3}{7}\right):\left(24-25+\frac{4}{21}-\frac{8}{21}\right)\)
\(\left(\frac{5.21-14+9}{21}\right):\left(\frac{-21-4}{21}\right)\)=\(\left(\frac{5.21-5}{21}\right).\left(\frac{21}{-25}\right)\)=\(\frac{5\left(21-1\right)}{\left(-5\right).5}=\frac{20}{-5}\)
=-4
#) Trả lời :
\(\frac{5^{2009}.5^3}{5^{2010}}=\frac{5^{2012}}{5^{2010}}=5^2=25\)
#Hok_tốt
a) \(S=1+2+2^2+...+2^{100}\)
\(2S=2+2^2+2^3+...+2^{101}\)
\(2S-S=\left(2+2^2+...+2^{101}\right)-\left(1+2+...+2^{100}\right)\)
\(S=2^{101}-1\)
b) \(X=2^{2012}-2^{2011}-...-2-1\)
\(X=2^{2012}-\left(1+2+...+2^{2011}\right)\)
Đặt \(X=2^{2012}-Y\)
Ta có :
\(Y=1+2+...+2^{2011}\)
\(2Y=2+2^2+...+2^{2012}\)
\(2Y-Y=\left(2+2^2+...+2^{2012}\right)-\left(1+2+...+2^{2011}\right)\)
\(Y=2^{2012}-1\)
\(\Rightarrow X=2^{2012}-2^{2012}+1\)
\(\Rightarrow X=1\)
\(\Rightarrow2010X=2010\)
a, ( 1/2 - 13/14 ) : 5/7 - ( -2/21 + 1/7 ) : 5/7
= ( 7/14 - 13/14 ) : 5/7 - ( -2/21 + 3/21 ) : 5/7
= -3/7 : 5/7 - 1/21 : 5/7
= -3/7 . 7/5 - 1/21 . 7/5
= ( -3/7 - 1/21 ) . 7/5
= ( -9/7+ -1/21 ) . 7/5
= -10/21 . 7/5
= -2/3
b, (68/5 + 19/4 ) - 43/5
= 68/5 + 19/4 - 43/5
= ( 68/5 - 43/5 ) + 19/4
= 5 +19/4
= 39/4
c, ( 93/11 + 29/8 ) - 38/11
= 93/11 + 29/8 - 38/11
= ( 93/11 - 38/11 ) + 29/8
= 5 + 29/8
=69/8
d, 4/9 : ( -1/7 ) + 59/9 : ( -1/7 )
= 4/9 . -7 + 59/9 . -7
= ( 4/9 + 59/9 ) . -7
= 7 . -7
= -49
