(7 ^2003 +7^2002) : 7^2001
a) 1 - 2 - 3 + 4 +5 - 6 - 7 + ..... + 2001 - 2002 -2003 + 2004
b) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ..... + 2001 + 2002 - 2003 - 2004
a) \(1-2-3+4+5-6-7+...+2001-2002-2003+2004\)
\(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(2001-2002-2003+2004\right)\)
\(=0+0+...+0=0\)
b) \(1+2-3-4+5+6-7-8+...+2001+2002-2003-2004\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(2001+2002-2003-2004\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=\left(-4\right)\cdot501=\left(-2004\right)\)
tinh nhanh
(7^2003+7^2002):7^2001
\(\left(7^{2003}+7^{2002}\right):7^{2001}\)
\(=\left(7^{2003}+7^{2002}\right).\frac{1}{7^{2001}}\)
\(=\frac{7^{2003}}{7^{2001}}+\frac{7^{2002}}{7^{2001}}\)
\(=7^2+7=49+7=56\)
2001*2002+2003*1991+11+2002*7
đáp số là ; 8008000
CHÚC BẠN HỌC TỐT !
(72003+72002)÷(72001×7)
Toan lop 6 kho the
minh lop 6 ne
tk nhe😉😉 tk lai cho😆😆😇😇
( 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
(7200372002)÷(72001×7)
\(\left(7^{2003}.7^{2002}\right):\left(7^{2001}.7\right)\)
\(=\left(7^{2003}.7^{2002}\right):\left(7^{2001+1}\right)\)
\(=7^{2003}.7^{2002}:7^{2002}=7^{2003}.\left(7^{2002}:7^{2002}\right)\)
\(=7^{2003}.1=7^{2003}\)
\(\left(7^{2003}+7^{2002}\right):\left(7^{2001}.7\right)\)
\(\left(7^{2003}+7^{2002}\right):\left(7^{2001}.7\right)\)
\(=\left(7^{2003}+7^{2002}\right):7^{2002}\)
\(=7^{2003}:7^{2002}+7^{2002}:7^{2002}\)
\(=7^{2003}:7^{2002}+1\)
\(=7^{2002}.7:7^{2002}.1+1\)
\(=7^{2002}.\left(7-1\right)+1\)
\(=7^{2002}.6+1\)
Ta có: \(\left(7^{2003}+7^{2002}\right):\left(7^{2001}.7\right)\)
\(\Rightarrow\left(7^{2003}+7^{2002}\right):7^{2002}\)
\(\Rightarrow7^{2002}:7^{2003}+7^{2002}:7^{2002}\)
Tự tính tiếp nha
=(7^2002*7+7^2002*1):(7^2002)
=7^2002*8:7^2002
=8
Tính
(7^2003+7^2002) chia het cho 7 ^ 2001
72003 chia het cho 72001
72002 chia het cho 72001
=> 72003 + 72002 chia het cho 7
=>
Ta có \(7^{2003}+7^{2002}\)
\(=7^{2001}.\left(7^2+7\right)\)
Ta thấy \(7^{2001}⋮7^{2001}\Rightarrow7^{2001}.\left(7^2+7\right)\)
Do đó \(7^{2003}+7^{2002}⋮7^{2001}\)
Vậy....
Tính :
7^2003 + 7^2002 chia hết cho 7^2001
7^2003 + 7^2002
= 7^2001 . 7^2 + 7^2001 . 7^1
= 7^2001 . 49 + 7^2001 . 7
= 7^2001 . ( 49 + 7 )
= 7^2001 . 56 chia hết cho 7^2001
Vậy 7^2003 + 7^2002 chia hết cho 7^2001 .
A) ( 7^2005 + 7^2004): 7^2001
B) ( 11^2003 + 11^2002): 7^2002
C) ( 5^2001 - 5^2000): 5 ^ 200
D) ( 7^2005 + 7^2004): 7^2001
Giải rõ dùm em vs ạ
Em đang cần gấp lắm nên mong mấy ac giúp vs
Tính:
( 72003 + 72002 ) + 72001