Tìm số tự nhiên x biết:
a) 3x-1 + 5 . 3x-1 = 162 ; b) 2x + 3 + 2x = 144
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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: =>3x-9+26 chia hết cho x-3
=>\(x-3\in\left\{1;-1;2;-2;13;-13;26;-26\right\}\)
=>\(x\in\left\{4;2;5;1;16;-10;29;-23\right\}\)
b: =>6x+38 chia hết cho 2x-3
=>6x-9+47 chia hết cho 2x-3
=>\(2x-3\in\left\{1;-1;47;-47\right\}\)
=>\(x\in\left\{2;1;25;-22\right\}\)
a) \(5\times x-123=12\)
\(\Rightarrow5\times x=135\)
\(\Rightarrow x=27\)
b) \(x+3x+5x+7x=96\)
\(\Rightarrow16x=96\)
\(\Rightarrow x=6\)
a) \(5\times x-123=12\)
\(5x=12+123\)
\(5x=135\)
\(x=135:5\)
\(x=27\)
________
b) \(x+3x+5x+7x=96\)
\(x\left(1+3+5+7\right)=96\)
\(x.16=96\)
\(x=96:16\)
\(x=6\)
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: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
a: \(\Leftrightarrow-x^2-3x+x+3+x^2-6x=11\)
=>-8x+3=11
=>-8x=8
hay x=-1
b: \(\Leftrightarrow3x^2-15x+x-5-3x^2+3x=5\)
=>-11x=10
hay x=-10/11