cmr \(7^{20}+49^{11}+343^7⋮57\)
CMR : \(7^{20}+49^{11}+343^7\) chia hết cho 57
\(7^{20}+49^{11}+343^7\)
\(=7^{20}+\left(7^2\right)^{11}+\left(7^3\right)^7\)
\(=7^{20}+7^{22}+7^{21}\)
\(=7^{20}\left(1+7^2+7\right)\)
\(=7^{20}.57⋮57\)
\(\Leftrightarrowđpcm\)
CMR : \(7^{20}+49^{11}+343^7\) \(⋮\) \(57\)
$ 7^{20} + 49^{11} + 343^{7} \\ = 7^{20} + (7^{2})^{11} + (7^{3})^{7} \\ = 7^{20} + 7^{22} + 7^{21} \\ = 7^{20}(1 + 49 + 7) \\ = 7^{20} . 57 \vdots 57 $
Chứng minh: 7^20+49^11+343^7 chia hết cho 57 ?
7^20 + 49^11 + 343^7 = ( 7^1 )^20 + ( 7^2 )^11 + ( 7^3 )^7
=7^20 + 7^21 + 7^22 = 7^20 ( 1 + 7 + 7^2 ) = 720.57 Vì 57 chia hết cho 57 nên 7^20 .57 chia hết cho 57 => 7^20 + 49^11 + 343^7 chia hết cho 57
Chứng minh 7^20 + 49^11 + 343^7 chia hết cho 57
720 + 4911 + 3437 = ( 71 )20 + ( 72 )11 + ( 73 )7 =720 + 721 + 722 = 720 ( 1 + 7 + 72 ) = 720.57
Vì 57 chia hết cho 57 nên 720 .57 chia hết cho 57
=> 720 + 4911 + 3437 chia hết cho 57 ( đpcm )
Ta có: 7^20 + 49^11 + 343^7 = 7^20 + (7^2)^11 + (7^3)^7 = 7^20 + 7^22 + 7^21 = 7^20(1 + 7 + 7^2) = 7^20.57 chia hết cho 57
=>ĐPCM
Chứng tỏ rằng :
\(7^{20}+49^{11}+343^7\) chia hết cho57
Ta có: 720+4911+3437= 720+(72)11+ (73)7
= 720+722+721=720.(1+7+72)=720.57 chia hết cho 57
\(\Rightarrow\)720+4911+3437 chia hết cho 57
1)Tìm số nguyên n lớn nhất sao cho n150 < 5225
2) Chứng minh 720 + 4911 + 3437 chia hết cho 57
3) So sánh :
a) 2515 và 810 nhân 330
(1 -1/7 -1/49-1/343 ) / ( 4- 4/7+ 4/49-4/343) = ?
Bài1 :Tìm x:
( 6 : 3/5 - 17/16 . 6/7 / 21/5 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 / 10/3 + 2/9 ) . x = 215/96
Ai nhanh mk K ạ.
\(\left(\dfrac{6:\dfrac{3}{5}-\dfrac{17}{16}.\dfrac{6}{7}}{\dfrac{21}{5}.\dfrac{10}{11}+\dfrac{57}{11}}-\dfrac{\left(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{15}\right).\dfrac{12}{49}}{\dfrac{10}{3}+\dfrac{2}{9}}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\left(\dfrac{\dfrac{509}{56}}{9}-\dfrac{\dfrac{7}{12}.\dfrac{12}{49}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{\dfrac{1}{7}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{9}{224}\right).x=\dfrac{215}{96}\)
\(\Rightarrow\dfrac{1955}{2016}.x=\dfrac{215}{96}\)
\(\Rightarrow x=\dfrac{215}{96}:\dfrac{1955}{2016}\)
\(\Rightarrow x=\dfrac{903}{391}\)
`[ 6 : 3/5 - 17/16 . 6/7 : 21/5 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 : 10/3 + 2/9 ] . x = 215/96`
`=>[ 6 . 5/3 - 17/16 . 6/7 . 5/21 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=>[10- 51/56 . 6/7 . 5/21 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> [10- 153/196 . 5/21 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> [10- 255/1372 . 10/11 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> [10- 1275/7546 + 57/11 - (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`
`=> (10- 1275/7546 + 57/11 - 7/12. 12/49 . 3/10 + 2/9 ) . x = 215/96`
`=> ( 10- 1275/7546 + 57/11 -343/600 . 3/10 + 2/9 ) . x = 215/96`
`=> ( 10- 1275/7546 + 57/11 -343/2000 + 2/9 ) . x = 215/96`
`=>15,06357671 . x= 215/96`
`=> x= 215/96: 15,06357671`
`=>x= 0,1486754027`
Đề có phải như vậy không nhỉ ?
*Tìm số tự nhiên n,biết :
1)8.2^n = 128
2) 121 . 11^n = 1331
3) 7^n :49=343
1)\(8.2^n=128\Rightarrow2^n=128:8\Rightarrow2^n=16\Rightarrow2^n=2^4\Rightarrow n=4\)
2)\(121.11^n=1331\Rightarrow11^n=1331:121\Rightarrow11^n=11\Rightarrow n=1\)
3)\(7^n:49=343\Rightarrow7^n:7^2=7^3\Rightarrow7^n=7^3.7^2\Rightarrow7^n=7^5\Rightarrow n=5\)
nhớ **** cho mình nhé