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19 tháng 9 2018

ko hieu :)))))))))))))))

19 tháng 9 2018

2D = \(2^3+2^4+...+2^{1998}\)

\(\Rightarrow2D-D=2^3+2^4+...+2^{1998}-2^2-2^3-...-2^{1997}\)

\(\Leftrightarrow D=2^{1998}-4\)

14 tháng 6 2017

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

21 tháng 10 2017

\(D=\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(D=\dfrac{1}{1999.2000}-\left(\dfrac{1}{1998.1999}+\dfrac{1}{1997.1998}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)\(D=\dfrac{1}{1999.2000}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{1997.1998}+\dfrac{1}{1998.1999}+\dfrac{1}{1999.2000}\right)\)

\(D=\dfrac{1}{1999.2000}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{1997}-\dfrac{1}{1998}+\dfrac{1}{1998}-\dfrac{1}{1999}+\dfrac{1}{1999}-\dfrac{1}{2000}\right)\)\(D=\dfrac{1}{1999.2000}-\dfrac{1999}{2000}\)

7 tháng 9 2016

a,(2016 + 2).1008:2=1017072

b,(2001+5).500:2=501500

c,

1 tháng 10 2016

501 500

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: \(=1997-\left[10\cdot8:8+8\right]=1997-11=1986\)

23 tháng 11 2017

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,\(205-\left[1200-\left(4^2-2.3\right)^3\right]:40\)

\(=205-200:40\)

\(=200\)

24 tháng 11 2017

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: \(F=5\dfrac{6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot2\dfrac{3}{a}\)

\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)+

\(=\dfrac{5a+6-4a-6}{ab}\)

\(=\dfrac{1}{b}\)

\(=\dfrac{1}{1997}\)

28 tháng 12 2022

\(1997-\left[10.\left(2^3-56\right):2^3+2^3\right].2005\)

\(=1997-\left[10.\left(8-56\right):8+8\right].2005\)

\(=1997-\left[10.\left(-48\right):8+8\right].2005\)

\(=1997-\left[\left(-60\right)+8\right].2005\)

\(=1997-\left(-52\right).2005\)

\(=1997-\left(-104260\right)=106257\)

28 tháng 12 2022

1997-[10.(23- 56): 23+23 ].2005

= 1997-[10.(- 56): 23+23 ].2005

= 1997-[10.(-48): 23+23 ].2005

= 1997-[10.(-48): 8 + 8 ].2005

= 1997-(-52).2005

= 106257