Tìm số tự nhiên x, biết:
a) 2 x : 4 = 3 2 ;
b) 3 x : 3 2 = 243 ;
c) 256 : 4 x = 4 2 .
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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
a)
\(\begin{array}{l}\left( {9x - {2^3}} \right):5 = 2\\9x - {2^3} = 2.5\\9x - 8 = 10\\9x = 18\\x = 2\end{array}\)
Vậy \(x = 2\)
b)
\(\begin{array}{l}\left[ {{3^4} - \left( {{8^2} + 14} \right):13} \right]x = {5^3} + {10^2}\\\left[ {81 - \left( {64 + 14} \right):13} \right]x = 125 + 100\\\left[ {81 - 78:13} \right]x = 125 + 100\\\left[ {81 - 6} \right]x = 225\\75x = 225\\x = 3\end{array}\)
Vậy \(x = 3\)
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) \(18-\left(2x+5\right)=9\)
\(2x+5=18-9\)
\(2x+5=9\)
\(2x=9-5\)
\(2x=4\)
\(x=2\)
a) \(18-\left(2x+5\right)=9\)
\(\Rightarrow2x+5=18-9=9\)
\(\Rightarrow2x=9-5=4\Rightarrow x=4:2=2\)
b) \(23x-4=32\Rightarrow23x=32+4=36\Rightarrow x=\dfrac{36}{23}\)
c) \(\left(3x+2\right)^2=64\)
\(\Rightarrow\left[{}\begin{matrix}3x+2=8\\3x+2=-8\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{10}{3}\end{matrix}\right.\)
d) \(x\left(2x-12\right)=0\Rightarrow6x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
`a, 2/3 +3/4 = (8+9)/12=17/12.`
`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.
`=> x=2.`
`b, 5/6-1/4=(20-6)/24=7/12`.
`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.
`=> x=1`.
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, \(\left[\left(4x+28\right).3+55\right]:5=35\)
\(=>\left(4x+28\right).3+55=35.5\)
\(=>\left(4x+28\right).3+55=175\)
\(=>\left(4x+28\right).3=175-55\)
\(=>\left(4x+28\right).3=120\)
\(=>4x+28=120:3\)
\(=>4x+28=40\)
\(=>4x=40-28\)
\(=>4x=12\)
\(=>x=12:4\)
\(=>x=3\)
\(#WendyDang\)
a) Ta có : 2 x : 2 2 = 2 5 nên x = 7.
b) Ta có: 3 x : 3 2 = 3 5 nên x = 7.
c) Ta có : 4 4 : 4 x = 4 2 nên x = 2.