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A = \(\dfrac{7^{2000}+1}{7^{2021}+1}\) ⇒ 7A = \(\dfrac{7^{2021}+7}{7^{2021}+1}\) = 1 + \(\dfrac{6}{7^{2021}+1}\)
B = \(\dfrac{7^{2021}+1}{7^{2022}+1}\) ⇒ 7B = \(\dfrac{7^{2022}+7}{7^{2022}+1}\) = 1 + \(\dfrac{6}{7^{2022}+1}\)
Vì \(\dfrac{6}{7^{2021}+1}\) > \(\dfrac{6}{7^{2022}+1}\) nên 7A > 7B (phân số nào có phần hơn lớn hơn thì phân số đó lớn hơn)
7A > 7B
A>B
\(\left(7^{2003}+7^{2002}\right):7^{2001}\)
\(=\left(7^{2003}+7^{2002}\right).\frac{1}{7^{2001}}\)
\(=\frac{7^{2003}}{7^{2001}}+\frac{7^{2002}}{7^{2001}}\)
\(=7^2+7=49+7=56\)
( 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
\(\left(7^{2003}.7^{2002}\right):\left(7^{2001}.7\right)\)
\(=\left(7^{2003}.7^{2002}\right):\left(7^{2001+1}\right)\)
\(=7^{2003}.7^{2002}:7^{2002}=7^{2003}.\left(7^{2002}:7^{2002}\right)\)
\(=7^{2003}.1=7^{2003}\)
\(\left(7^{2003}+7^{2002}\right):\left(7^{2001}.7\right)\)
\(=\left(7^{2003}+7^{2002}\right):7^{2002}\)
\(=7^{2003}:7^{2002}+7^{2002}:7^{2002}\)
\(=7^{2003}:7^{2002}+1\)
\(=7^{2002}.7:7^{2002}.1+1\)
\(=7^{2002}.\left(7-1\right)+1\)
\(=7^{2002}.6+1\)
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
= ( 1 + 7) + ( 72 + 73 ) + ... + ( 72000 + 72001)
= ( 1 + 7) + 72( 1 + 7) + ... + 72000( 1 + 7)
= ( 1 + 7) x ( 1 + 72 + ... + 72000) : hết cho 8
\(=\dfrac{7^{2073}}{7^{2001}}+\dfrac{7^{2072}}{7^{2001}}=7^{72}+7^{71}\)