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\(\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).\left(2^4-4^2\right)\)
\(=\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).\left(16-16\right)\)
\(=\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).0\)
\(=0\)
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\(\left(7^{1997}-7^{1995}\right):\left(7^{1994}.7\right)\)
\(=\left[7^{1995}\left(7^2-1\right)\right]:7^{1995}\)
\(=7^{1995}.48:7^{1995}\)
\(=48\)
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\(\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(3^8-81^2\right)\)
\(=\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(6561-6561\right)\)
\(=\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).0\)
\(=0\)
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\(\left(2^8+8^3\right):\left(2^5.2^3\right)\)
\(=\left[2^8+\left(2^3\right)^3\right]:2^8\)
\(=\left(2^8+2^9\right):2^8\)
\(=2^8.\left(1+2\right):2^8\)
\(=2^8.3:2^8\)
\(=3\)
tính
b)
c)
tìm x
b)
c)
d) (với x là số nguyên dương)
Bài 1:
b: \(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}=\dfrac{1}{2}\)
c: \(=\dfrac{1}{9}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)-\dfrac{8}{9}=\dfrac{1}{9}-\dfrac{8}{9}=-\dfrac{7}{9}\)
Bài 2:
b: \(\Leftrightarrow x+\dfrac{3}{10}=\dfrac{17}{30}\)
hay x=4/15
2008 . 2007 . 2007 - 2007 . 2008 - 2008
= ( 2007 + 1 ) . 2007 . 2007 - 2007 . ( 2007 + 1 ) - 2008
= 2007 . 2007 . 2007 + 2007 . 2007 - 2007 . 2007 + 2007 - 2008
= 20073 + 0 +2007 - 2008
= 20073 + 2007 - 2008
= 20073 + -1
Mình trả lời dc rồi cám ơn bạn nhưng kết quả của bạn sai rồi kết quả đúng là 0 nha
\(B=\left(4^5\cdot10\cdot5^6+25^5\cdot2^8\right):\left(2^8\cdot5^4+5^7\cdot2^5\right)\)
\(B=\dfrac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{2^8\cdot5^7\cdot\left(2^3\cdot1+5^3\cdot1\right)}{2^5\cdot5^4\cdot\left(2^3\cdot1+5^3\cdot1\right)}\)
\(B=\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\)
\(B=\dfrac{2^3\cdot5^3}{1\cdot1}\)
\(B=\left(2\cdot5\right)^3\)
\(B=10^3\)
\(B=1000\)
Đặt A = 1 + 2 + 22 + ... + 22008
2A = 2 + 22 + 23 + ... + 22009
2A - A = A = 1 - 22009
=> B = 1 - 22009 / 1 - 22009
=> B = 1
Tk mk nha!
Đặt C\(=1+2+2^2+....+2^{2008}\)
2C\(=2+2^2+2^3+....+2^{2009}\)
2C-C\(=2^{2009}-1\)
C\(=2^{2009}-1\)
Vậy B=\(\frac{2^{2009}-1}{1-2^{2009}}\)
\(B=\left(17+\dfrac{1}{25}-16-\dfrac{11}{15}\right)\cdot\dfrac{75}{23}+\dfrac{75}{23}\cdot\dfrac{23}{75}\)
\(=\dfrac{23}{75}\cdot\dfrac{75}{23}+1=1+1=2\)
(71997-71995):(71994.7) = (71997-71995):71995 = 71997 : 71995 - 71995 : 71995 = 72 : 1 = 49
(71997-71995):(71994.7)
=(71997-71995): 71995
=71997:71995-71995:71995
=72:1
=49:1=49
CHÚC BẠN HỌC GIỎI!!