Tìm số tự nhiên x , biết:
3) x \(⋮\) 25 ; x \(⋮\) 30 và 500 < x < 1000
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a) 2011 : { 639 : [ 316 – ( 78 + 25 )] : 3 }
= 2011 : { 639 : [ 316 – 103 ] : 3}
= 2011 : ( 639 : 213 : 3 ) = 2011 : (3 : 3 ) = 2011 : 1 = 2011
b) ( 3x – 2 3 ) . 7 = 7 4
3x – 8 = 7 4 : 7
3x – 8 = 7 3
3x – 8 = 343
3x = 343 + 8
3x = 351
x = 351 : 3 = 117
c) (8705 + 5235) – 5x = 3885
13940 – 5x = 3885
5x = 13940 – 3885
5x = 10055
x = 10055 : 5 = 2011
\(Ư\left(30\right)=\left\{1;2;3;5;6;10;15;30\right\}\\ \Rightarrow x=5\left(B\right)\\ B\left(8\right)=\left\{0;8;16;24;32;...\right\}\\ \Rightarrow x=24\left(B\right)\)
a) 2011 : { 639 : [ 316 – ( 78 + 25 )] : 3 }
= 2011 : { 639 : [ 316 – 103 ] : 3}
= 2011 : ( 639 : 213 : 3 ) = 2011 : (3 : 3 ) = 2011 : 1 = 2011
b) ( 3x – 23) . 7 = 74
3x – 8 = 74 : 7
3x – 8 = 73
3x – 8 = 343
3x = 343 + 8
3x = 351
x = 351 : 3 = 117
c) (8705 + 5235) – 5x = 3885
13940 – 5x = 3885
5x = 13940 – 3885
5x = 10055
x = 10055 : 5 = 2011
Trl:
a) \(2011:\left\{639:\left[316-\left(78+25\right)\right]:3\right\}\)
\(=2011:\left\{639:\left[316-103\right]:3\right\}\)
\(=2011:\left\{639:213:3\right\}\)
\(=2011:1\)
\(=2011\)
b) \(\left(3x-23\right).7=74\)
\(\Rightarrow3x-23=74:7\)
\(\Rightarrow3x-23=10,5\)
\(\Rightarrow3x=10,5+23\)
\(\Rightarrow3x=33,5\)
\(\Rightarrow x=33,5:3\)
\(\Rightarrow11,1\)( Câu này sai đề nha )
c) \(\left(8705+5235\right)-5x=3885\)
\(\Rightarrow13940-5x=3885\)
\(\Rightarrow5x=10055\)
\(\Rightarrow x=10055:5\)
\(\Rightarrow x=2011\)
`(1/(1.3)+1/(3.5)+.......+1/(23.25))xx((x+1)+(x+3)+(x+5)+.....+(x+23))=144`
`(2/(1.3)+2/(3.5)+.......+2/(23.25))xx[(x+x+....+x)+(1+3+5+...+23)]=288`
`(1-1/3+1/3-1/5+.....+1/23-1/25)xx(12x+(24.12)/2)=288`
`(1-1/25)xx(12x+12.12)=288`
`24/25xx[12(x+12)]=288`
`24/25xx(x+12)=28`
`x+12=28:24/25=50`
`x=50-12=38`
Vậy `x=38`
a) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)-10=40\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow x+7=\dfrac{50}{5}\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
b) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x-18=81\)
\(\Rightarrow9x=81+18\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
c) \(5^{25}\cdot5^{x-1}=5^{25}\)
\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)
\(\Rightarrow5^{x-1}=1\)
\(\Rightarrow5^{x-1}=5^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
a) 5(�+7)−10=23⋅55(x+7)−10=23⋅5
⇒5(�+7)−10=40⇒5(x+7)−10=40
⇒5(�+7)=40+10⇒5(x+7)=40+10
⇒�+7=505⇒x+7=550
⇒�+7=10⇒x+7=10
⇒�=10−7⇒x=10−7
⇒�=3⇒x=3
b) 9�−2⋅32=349x−2⋅32=34
⇒9�−18=81⇒9x−18=81
⇒9�=81+18⇒9x=81+18
⇒9�=99⇒9x=99
⇒�=999⇒x=999
⇒�=11⇒x=11
c) 525⋅5�−1=525525⋅5x−1=525
⇒5�−1=525:525⇒5x−1=525:525
⇒5�−1=1⇒5x−1=1
⇒5�−1=50⇒5x−1=50
⇒�−1=0⇒x−1=0
⇒�=1⇒x=1
Ta có :
\(\begin{cases}x⋮25\\x⋮30\end{cases}\)\(\Rightarrow x\inƯC_{\left(25;30\right)}\)
Mà ƯCLN(25;30)=300
=> \(x\in\left\{0;300;600;900;...\right\}\)
Mà 500 < x < 1000
=> x = 600 ; x = 900
Vậy x = 600 ; x = 900
Vì x \(⋮\)25;x\(⋮\)30=>xϵBC(25;30)
ta có:
25=52
30=2.3.5
=>BCNN(25;30)=2.3.52=150
=>BC(25;30)={0;150;300;450;600;750;900;1050.....}
Mà 500<x<1000=>xϵ{600;750;900}