D=2^2+2^3+...+2^1997
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a) A=1-2-3+4+5-6-7+.....+1996+1997-1998-1999+2000
=(1-2-3+4)+(5-6-7+8)+...+(1997-1998-1999+2000)
=0
b) B=1-3+5-7+....+2001-2003+2005
=(1-3)+(5-7)+...+(2001-2003)+2005
=-2.501+2005
=-1002+2005
=1003
c) C=1-2-3+4+5-6-7+8+.....+1993-1994-1995+1996+1997
=(1-2-3+4)+(5-6-7+8)+...+(1993-1994-1995+1996)+1997
=1997
d) D=1000+998+996+......+10-999-997-995-...-11
=(1000-999)+(998-997)+(996-995)+....+(12-11)+10
=1.495+10
=595
\(D=\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(D=\dfrac{1}{1999.2000}-\left(\dfrac{1}{1998.1999}+\dfrac{1}{1997.1998}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)\(D=\dfrac{1}{1999.2000}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{1997.1998}+\dfrac{1}{1998.1999}+\dfrac{1}{1999.2000}\right)\)
\(D=\dfrac{1}{1999.2000}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{1997}-\dfrac{1}{1998}+\dfrac{1}{1998}-\dfrac{1}{1999}+\dfrac{1}{1999}-\dfrac{1}{2000}\right)\)\(D=\dfrac{1}{1999.2000}-\dfrac{1999}{2000}\)
Đặt 4453=a; 1997=b
\(A=\left(5+\dfrac{6}{a}\right)\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\left(2+\dfrac{3}{a}\right)\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)
\(=\dfrac{5a+6-4a-6}{ab}=\dfrac{a}{ab}=\dfrac{1}{b}=\dfrac{1}{1997}\)
d: =-452+67-75+452=-8
e: \(=1997-\left[10\cdot8:8+8\right]=1997-11=1986\)
2.
a,\(50-\left[\left(50-2^3.5\right):2+3\right]\)
\(=50-\left[\left(50-40\right):2+3\right]\)
\(=50-\left(10:2+3\right)\)
\(=50-8\)
\(=42\)
b,\(8697-\left[3^7:3^5+2\left(13-3\right)\right]\)
\(=8697-\left(3^2+2.10\right)\)
\(=8697-\left(9+20\right)\)
\(=8697-29\)
\(=8668\)
c,\(205-\left[1200-\left(4^2-2.3\right)^3\right]:40\)
\(=205-200:40\)
\(=200\)
2)
a) \(50-\left[\left(50-2^3.5\right):2+3\right]\)
\(=50-\left[\left(50-8.5\right):2+3\right]\)
\(=50-\left[\left(50-40\right):2+3\right]\)
\(=50-\left(10:2+3\right)\)
\(=50-\left(5+3\right)\)
\(=50-8\)
\(=42\)
b) \(8697-\left[3^7:3^5+2\left(13-3\right)\right]\)
\(=8697-\left(3^7:3^5+2.10\right)\)
\(=8697-\left(3^{7-5}+2.10\right)\)
\(=8697-\left(3^2+2.10\right)\)
\(=8697-\left(9+2.10\right)\)
\(=8697-\left(9+20\right)\)
\(=8697-29\)
\(=8668\)
c) \(205-\left[1200-\left(4^2-2.3\right)^3\right]:40\)
\(=205-\left[1200-\left(16-2.3\right)^3\right]:40\)
\(=205-\left[1200-\left(16-6\right)^3\right]:40\)
\(=205-\left(1200-10^3\right):40\)
\(=205-\left(1200-1000\right):40\)
\(=205-200:40\)
\(=205-5\)
\(=200\)
Đặt a=4453, b=1997
Ta có: \(F=5\dfrac{6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot2\dfrac{3}{a}\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)+
\(=\dfrac{5a+6-4a-6}{ab}\)
\(=\dfrac{1}{b}\)
\(=\dfrac{1}{1997}\)
\(1997-\left[10.\left(2^3-56\right):2^3+2^3\right].2005\)
\(=1997-\left[10.\left(8-56\right):8+8\right].2005\)
\(=1997-\left[10.\left(-48\right):8+8\right].2005\)
\(=1997-\left[\left(-60\right)+8\right].2005\)
\(=1997-\left(-52\right).2005\)
\(=1997-\left(-104260\right)=106257\)
1997-[10.(23- 56): 23+23 ].2005
= 1997-[10.(8 - 56): 23+23 ].2005
= 1997-[10.(-48): 23+23 ].2005
= 1997-[10.(-48): 8 + 8 ].2005
= 1997-(-52).2005
= 106257
ko hieu :)))))))))))))))
2D = \(2^3+2^4+...+2^{1998}\)
\(\Rightarrow2D-D=2^3+2^4+...+2^{1998}-2^2-2^3-...-2^{1997}\)
\(\Leftrightarrow D=2^{1998}-4\)