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7^2003/7^2001 +7^2002/7^2001
=7^2+7=49+7
=56
Neu dung nho chon cay tra loi cua to nhe
\(=7^{2003}:7^{2001}+7^{2002}:7^{2001}=7^{2003-2001}+7^{2002-2001}\)
\(=7^2+7^1=49+7=56\)
M=70+71+72+...+72018+72019
M=(70+71)+(72+74)+...+(72018+72019)
M=70.(70+71)+72.(70+71)+.....+72018.(70+71)
=70.8+72.8+...+72018.8
=8.(70+72+...+72018) chia hết cho 8
=>M là bội của 8
Chúc bn học tốt
\(a,\left(7^{2003}+7^{2002}\right):7^{2001}\)
\(=7^{2003}:7^{2001}+7^{2002}:7^{2001}\)
\(=7^2+7\)
\(=49+7\)
\(=56\)
a, \(\left(7^{2003}+7^{2002}\right)\div7^{2001}=7^{2001}.\left(7^2+7\right)\div7^{2001}=7^2+7=56\)
b, \(\left(5^4+4^7\right)\left(8^9-2^7\right)\left(2^4-4^2\right)\)
\(=\left(5^4+4^7\right)\left(8^9-2^7\right)\left(16-16\right)\)
\(=\left(5^4+4^7\right)\left(8^9-2^7\right)0=0\)
=7^2003:7^2000+7^2002:7^2000+7^2001:7^2000
=7^3+7^2+7
=343+49+7=399
Từ gt suy ra: -4C= 4101+...+44+43-42
=> -4C+C= 4101-42-42+4=4101-28 => C=\(\frac{28-4^{101}}{3}\)
( 72003 + 72002 ) \(\div\)( 72001 x 7 )
= ( 72003 + 72002 ) \(\div\)( 72001+1)
= ( 72003 + 72002 ) \(\div\)72002
= ( 72003 \(\div\) 72002 ) + ( 72002 \(\div\)72002)
= 72003-2002 + 72002-2002
= 71 + 70
= 7 + 1 = 8
HK TỐT
\(\left(7^{2003}.7^{2002}\right):\left(7^{2001}.7\right)\)
\(=\left(7^{2003}.7^{2002}\right):\left(7^{2001+1}\right)\)
\(=7^{2003}.7^{2002}:7^{2002}=7^{2003}.\left(7^{2002}:7^{2002}\right)\)
\(=7^{2003}.1=7^{2003}\)
A) \(\frac{7^{2003}+7^{2002}}{7^{2004}}=\frac{7^{2003}}{7^{2004}}+\frac{7^{2002}}{7^{2004}}=\frac{1}{7}+\frac{1}{7^2}=\frac{8}{49}\)
B) Xem lại đề bài
\(\frac{7^{2003}+7^{2002}}{7^{2001}}\)=\(\frac{7^{2003}}{7^{2001}}+\frac{7^{2002}}{7^{2001}}\)=\(\frac{7^{2001}.7^2}{7^{2001}}+\frac{7^{2001}.7}{7^{2001}}\)= 72 + 7 = 56