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3x+ 10 chia hết cho x + 1
3x + 3 + 7 chia hết cho x + 1
7 chia hết cho x + 1
x+ 1 thuộc U(7) = {1;7}
x + 1 = 1 => x = 0
x + 1 = 7 => x= 6
a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)
b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)
\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)
c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)
\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)
e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)
f) \(\left|12-x\right|-7=5\)
th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)
th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)
i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)
k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)
a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)
b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)
Làm nốt nha
a, \(|x+10|=0\) b,\(|x|< 10\)=> \(-10< x< 10\)
c,\(|x-1|=5\)
=>\(x+10=0\) => \(\hept{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)=>\(\hept{\begin{cases}x=6\\x=-4\end{cases}}\)
=>\(x=-10\) d, (x + 2) . (x - 8 ) = 0
\(\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\)=>\(\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)
\(\left(3x-1\right)⋮\left(x+1\right)\)
\(\Rightarrow\left(3x+3-4\right)⋮\left(x+1\right)\)
\(\Rightarrow\left(-4\right)⋮\left(x+1\right)\)
\(\Rightarrow x+1\inƯ\left(-4\right)=\left\{-4;-1;1;4\right\}\)
\(\Rightarrow x\in\left\{-5;-2;0;3\right\}\)
a) 10 ⋮ (x - 1)
⇒ x - 1 ∈ Ư(10) = {-10; -5; -2; -1; 1; 2; 5; 10}
⇒ x ∈ {-9; -4; -1; 0; 2; 3; 6; 11}
b) x + 5 = x - 2 + 7
Để (x - 5) ⋮ (x - 2) thì 7 ⋮ (x - 2)
⇒ x - 2 ∈ Ư(7) = {-7; -1; 1; 7}
⇒ x ∈ {-5; 1; 3; 9}
c) 3x + 8 = 3x - 3 + 11
= 3(x - 1) + 11
Để (3x + 8) ⋮ (x - 3) thì 11 ⋮ (x - 3)
⇒ x - 3 ∈ Ư(11) = {-11; -1; 1; 11}
⇒ x ∈ {-8; 2; 4; 14}