Tìm số tự nhiên x biết:
a. 3 x = 1
b. x 4 = 1
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a: =>3x+3=5x-25
=>-2x=-28
hay x=14
b: =>3x+6=-4x+20
=>7x=14
hay x=2
`a,`\(2^x -15= 2^4+1\)
`-> 2^x-15=17`
`-> 2^x=17+15`
`-> 2^x=32`
`-> 2^x=2^5`
`-> x=5`
`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?
`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)
`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)
`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
Mà `1/65+1/64-1/63-1/62 \ne 0`
`-> x+66=0`
`-> x=-66`
a: =>2^x=2^4+16=32
=>x=5
b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)
=>x+66=0
=>x=-66
Bài 10:
a: 2x-3 là bội của x+1
=>\(2x-3⋮x+1\)
=>\(2x+2-5⋮x+1\)
=>\(-5⋮x+1\)
=>\(x+1\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{0;-2;4;-6\right\}\)
b: x-2 là ước của 3x-2
=>\(3x-2⋮x-2\)
=>\(3x-6+4⋮x-2\)
=>\(4⋮x-2\)
=>\(x-2\inƯ\left(4\right)\)
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6;-2\right\}\)
Bài 14:
a: \(4n-5⋮2n-1\)
=>\(4n-2-3⋮2n-1\)
=>\(-3⋮2n-1\)
=>\(2n-1\inƯ\left(-3\right)\)
=>\(2n-1\in\left\{1;-1;3;-3\right\}\)
=>\(2n\in\left\{2;0;4;-2\right\}\)
=>\(n\in\left\{1;0;2;-1\right\}\)
mà n>=0
nên \(n\in\left\{1;0;2\right\}\)
b: \(n^2+3n+1⋮n+1\)
=>\(n^2+n+2n+2-1⋮n+1\)
=>\(n\left(n+1\right)+2\left(n+1\right)-1⋮n+1\)
=>\(-1⋮n+1\)
=>\(n+1\in\left\{1;-1\right\}\)
=>\(n\in\left\{0;-2\right\}\)
mà n là số tự nhiên
nên n=0
B3:
a) x = 0
b) 9
B4 :
a) x = 1,2,3
b) x = 0,1,2
B5:
0,21; 0,215; 0,22
a) \(128-3\left(x+4\right)=23\)
\(\Rightarrow3\left(x+4\right)=128-23\)
\(\Rightarrow3\left(x+4\right)=105\)
\(\Rightarrow x+4=35\)
\(\Rightarrow x=35-4\)
\(\Rightarrow x=31\)
b) \(\left[\left(4x+28\right)\cdot3+55\right]:5=35\)
\(\Rightarrow\left(4x+28\right)\cdot3+55=35\cdot5\)
\(\Rightarrow\left(4x+28\right)\cdot3+55=175\)
\(\Rightarrow\left(4x+28\right)\cdot3=120\)
\(\Rightarrow4x+28=40\)
\(\Rightarrow4x=12\)
\(\Rightarrow x=3\)
a, \(128-3\left(x+4\right)=23\)
\(=>3\left(x+4\right)=128-23\)
\(=>3\left(x+4\right)=105\)
\(=>x+4=105:3\)
\(=>x+4=35\)
\(=>x=35-4\)
\(=>x=31\)
b, \(\left[\left(4x+28\right).3+55\right]:5=35\)
\(=>\left(4x+28\right).3+55=35.5\)
\(=>\left(4x+28\right).3+55=175\)
\(=>\left(4x+28\right).3=175-55\)
\(=>\left(4x+28\right).3=120\)
\(=>4x+28=120:3\)
\(=>4x+28=40\)
\(=>4x=40-28\)
\(=>4x=12\)
\(=>x=12:4\)
\(=>x=3\)
\(#WendyDang\)
Tham khảo:
a)
( 2x + 1 ) . ( y - 3 ) = 12
Vì 2x +1 là số lẻ.
Do ( 2x + 1 ) . ( y - 3) = 12
=> 2x + 1 : y - 3 thuộc Ư ( 12) = { 1 ; 2 ; 3 ; 4 ; 6 ; 12 }
=> 2 x +1 = 1 => x= 0
hoặc y - 3 = 12 => y = 15
=> 2x + 1 = 3 => x = 2
hoặc y - 3 = 4 => y = 7
=> 2x + 1 = 2 ( L)
VẬY ( x ; y) = { ( 0 ; 15 ) ; ( 2 ; 7) }
Bài 1:
a: Ta có: \(48751-\left(10425+y\right)=3828:12\)
\(\Leftrightarrow y+10425=48751-319=48432\)
hay y=38007
b: Ta có: \(\left(2367-y\right)-\left(2^{10}-7\right)=15^2-20\)
\(\Leftrightarrow2367-y=1222\)
hay y=1145
Bài 2:
Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow288:\left(x-3\right)^2=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
a. 3 x =1= 3 0 ⇒ x = 0
b. x 4 =1= 1 4 ⇒x = 1