1005 + 7458 * 1005 + 1487
tính bằng cách " lề mề" nhất
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BT
1
AH
Akai Haruma
Giáo viên
12 tháng 5 2022
Lời giải:
$2010\times 4+1005\times 190+2019$
$=1005\times 2\times 4+1005\times 190+2019$
$=1005\times 8+1005\times 190+2019$
$=1005\times (8+190)+2019=1005\times 198+2019=201009$
LH
1
17 tháng 4 2018
\(\frac{2010.2011-1005}{2010.2010+1005}\)
\(=\frac{2010.\left(2010+1\right)-2005}{2010.2010+1005}\)
\(=\frac{2010.2010+2010-1005}{2010.2010+1005}\)
\(=\frac{2010.2010+1005}{2010.2010+1005}\)
\(=1\)
8 tháng 9 2017
https://olm.vn/hoi-dap/question/1038454.html
Mình vừa làm cách đây 11 phút nhé !
8 tháng 9 2017
Ta có : a2010 + b2010 + c2010 = a1005b1005 + b1005c1005 + c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 = 2a1005b1005 + 2b1005c1005 + 2c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 - 2a1005b1005 - 2b1005c1005 - 2c1005a1005 = 0
<=> (a2010 - 2a1005b1005 + b2010) + (b2010 - 2b1005c1005 + c2010) + (c2010 - 2c1005a1005 + a2010) = 0
<=> (a1005 - b1005)2 + (b1005 - c1005)2 + (c1005 - a1005 )2 = 0
=> a1005 - b1005 = b1005 - c1005 = c1005 - a1005 = 0
=> a = b = c
Vậy (a - b)20 + (b - c)11 + (c - a)2010 = (a - a)20 + (a - a)11 + (a - a)2010 = 0 + 0 + 0 = 0 .
8 tháng 9 2017
Ta có : a2010 + b2010 + c2010 = a1005b1005 + b1005c1005 + c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 = 2a1005b1005 + 2b1005c1005 + 2c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 - 2a1005b1005 - 2b1005c1005 - 2c1005a1005 = 0
<=> (a2010 - 2a1005b1005 + b2010) + (b2010 - 2b1005c1005 + c2010) + (c2010 - 2c1005a1005 + a2010) = 0
<=> (a1005 - b1005)2 + (b1005 - c1005)2 + (c1005 - a1005 )2 = 0
=> a1005 - b1005 = b1005 - c1005 = c1005 - a1005 = 0
=> a = b = c
Vậy (a - b)20 + (b - c)11 + (c - a)2010 = (a - a)20 + (a - a)11 + (a - a)2010 = 0 + 0 + 0 = 0 .
AT
19 tháng 2 2018
a2010 + b2010 + c2010 = a1005b1005 + b1005c1005 + c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 = 2a1005b1005 + 2b1005c1005 + 2c1005a1005
<=> 2a2010 + 2b2010 + 2c2010 - 2a1005b1005 - 2b1005c1005 - 2c1005a1005 = 0
<=> (a2010 - 2a1005b1005 + b2010) + (b2010 - 2b1005c1005 + c2010) + (c2010 - 2c1005a1005 + a2010) = 0
<=> (a1005 - b1005)2 + (b1005 - c1005)2 + (c1005 - a1005 )2 = 0
=> a1005 - b1005 = b1005 - c1005 = c1005 - a1005 = 0
=> a = b = c
LT
1
I
25 tháng 2 2020
ta có \(\frac{a}{b}=\frac{c}{d}\)
=>\(\frac{a}{c}=\frac{b}{d}\)(1)
Từ (1) => \(\frac{a^{1005}}{c^{1005}}=\frac{b^{1005}}{d^{1005}}=\frac{a^{1005}+b^{1005}}{c^{1005}+d^{1005}}\)(2)
Từ (1) => \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
=>\(\left(\frac{a}{c}\right)^{1005}=\left(\frac{b}{d}\right)^{1005}=\left(\frac{a+b}{c+d}\right)^{1005}=\frac{\left(a+b\right)^{1005}}{\left(c+d\right)^{1005}}\)(3)
mà \(\left(\frac{a}{c}\right)^{1005}=\frac{a^{1005}}{c^{1005}}\)
từ 2 zà 3 => ghi lại cái cần chứng minh nha ( dpcm)
NT
0
Y
18 tháng 5 2016
\(\Leftrightarrow2\left(a^{2010}+b^{2010}+c^{2010}\right)=2\left(a^{1005}b^{1005}+b^{1005}c^{1005}+c^{1005}a^{1005}\right)\)
\(\Leftrightarrow2a^{2010}+2b^{2010}+2c^{2010}-2a^{1005}b^{1005}-2b^{1005}c^{1005}-2c^{1005}a^{1005}=0\)
\(\Leftrightarrow\left(a^{2010}-2a^{1005}b^{1005}+b^{2010}\right)+\left(b^{2010}-2b^{1005}c^{1005}+c^{2010}\right)+\left(c^{2010}-2c^{1005}a^{1005}+a^{2010}\right)=0\)
\(\Leftrightarrow\left(a^{1005}-b^{1005}\right)^2+\left(b^{1005}-c^{1005}\right)^2+\left(c^{1005}-a^{1005}\right)^2=0\)
\(\Rightarrow\left(a^{1005}-b^{1005}\right)^2=0;\left(b^{1005}-c^{1005}\right)^2=0;\left(c^{1005}-a^{1005}\right)^2=0\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\left(a-a\right)^{20}+\left(a-a\right)^{11}+\left(a-a\right)^{2010}=0\)
18 tháng 5 2016
2 ( a trên 2010 + b trân 2010 + c trên 2010 ) = 2 ( a trên 1005 b trên 1005 + b trên 1005 c trên 1005 + c trên 1005 a trên 1005 )
2a^ ( 2010 ) + 2b^ ( 2010 ) + 2c^ ( 2010 ) - 2a^ ( 1005 ) b^ ( 1005 ) - 2b^ ( 1005 ) c^ ( 1005 ) - 2c^ ( 1005 )a^ ( 1005 ) = O\)
( a^ ( 2010 ) - 2a^ ( 1005 ) b^ ( 1005 ) + b^ ( 2010 ) + ( b^( 2010 ) - 2b^ ( 1005 ) c^ ( 1005 ) + c^ ( 2010 ) + ( c^ ( 2010 ) - 2c^ ( 1005 ) a^ ( 1005 ) + a^ ( 2010 ) = 0\)
( a^ ( 1005 ) ^2 + ( b^ ( 1005 ) - c^ ( 1005 ) ^2 + ( c^ ( 1005 ) - a^ ( 1005 ) - a^ ( 1005 ) ^2 = 0\)
( a^ ( 1005 ) - b^ ( 1005 ) ^ 2= 0 : ( b^ ( 1005 ) - c^ ( 1005 ) ^2 = 0 : ( c^ ( 1005 ) - a^ ( 1005 ) ^2 = 0\)
a = b = c
( a - a ) ^ ( 20 ) + ( a - a ) ^ ( 11 ) + ( a - a ) ^ (2010 = 0\)
Vậy : ( a -a ) ^ ( 20 ) + ( a - a ) ^ ( 11 ) + ( a + a ) ^ ( 2010 = 0\)
ai cho tick de minh duoc **** di