Bài 1: Tính giá trị theo biểu thức.
a) 1000 x 10 + 2000 x 10 + 10000 x 10
b) 1001 x 10 + 2010 x 10 + 20110 x 10
c) 2000 x 10 + 4000 x 10 + 60000 x 10
d) 3000 x 10 + 2000 x 10 + 7000 x 10 + 2000 x 10
(Chỉ tính nhanh trong vòng 1 phút 10 giây)
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a ) 18 x 10 = 180
18 x 100 = 1800
18 x 1000 = 18000
82 x 100 = 8200
75 x 1000 = 75000
19 x 10 = 190
256 x 1000 = 256000
302 x 10 = 3020
400 x 100 = 40000
b ) 9000 : 10 = 900
9000 : 100 = 90
9000 : 1000 = 9
6800 : 100 = 68
420 : 10 = 42
2000 : 1000 = 2
20020 : 10 = 2002
200200 : 100 = 2002
2002000 : 1000 = 2002
a)18 x 10 = 180
18 x 100 = 1 800
18 x 1000 = 18 000
82 x 100 = 8 200
75 x 1000 = 75 000
19 x 10 = 190
256 x 1000 = 256 000
302 x 10 = 3 020
400 x 100 = 40 000
b)9000 : 10 = 900
9000 : 100 = 90
9000 : 1000 = 9
6800 : 100 = 68
420 : 10 = 42
2000 : 1000 = 2
20 020 : 10 = 2 002
200 200 : 100 = 2 002
2 002 000 : 1000 = 2 002
1000+2000+3000+x=10000
x=10000-1000-2000-3000
x=4000
vậy x = 4000
`a, 1000 : 100 = 10`
`b, 75 xx 100 = 7500`
`c, 900 xx 10 = 9000`
`d, 8 xx 100 : 10 : 10 xx 10000 = 80000`
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\(\dfrac{1}{\left(x+2000\right)\left(x+2001\right)}+\dfrac{1}{\left(x+2001\right)\left(x+2002\right)}+...+\dfrac{1}{\left(x+2009\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2001}+\dfrac{1}{x+2001}-\dfrac{1}{x+2002}+...+\dfrac{1}{x+2009}-\dfrac{1}{x+2010}=\dfrac{10}{11}\)
\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{x+2010-x-2000}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\)
\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{10}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\left(x+2000\right)\left(x+2010\right)=11\\ \Leftrightarrow...\)
(X -10/1994 -1) + (X-8/1996 - 1) + (X-6/1998 - 1)+ (X-4/2000 - 1) + (X-2/2002 - 1) = (X-2002/2 - 1) + (X-2000/4 - 1) + (X-1998/6 - 1) + (X-1996/8 - 1) + (X-1994/10 - 1)
=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 = x-2004/2 + x-2004/4 + x-2004/6 + x-2004/8 + x-2004/1994
=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 - x-2004/2 - x-2004/4 - x-2004/6 - x-2004/8 - x-2004/1994 = 0
=> (x - 2004)(1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 + 1/4 + 1/6 + 1/8) = 0
Mà (1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 + 1/4 + 1/6 + 1/8) \(\ne\)0
=> x - 2004 = 0
=> x = 2004
Vậy x = 2004
a, 130000
b, 231210
c, 660000
d, 140000
a, 130000
b, 231210
c, 660000
d, 140000