Tìm x
14 chia hết cho (2x - 1)
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Bài 1:
Ta có: \(5x^3-3x^2+2x+a⋮x+1\)
\(\Leftrightarrow5x^3+5x^2-8x^2-8x+10x+10+a-10⋮x+1\)
\(\Leftrightarrow a-10=0\)
hay a=10
a) Ta có: \(2x-2\)\(⋮\)\(x-2\)
\(\Leftrightarrow\)\(2\left(x-2\right)+2\)\(⋮\)\(x-2\)
Ta thấy \(2\left(x-2\right)\)\(⋮\)\(x-2\)
nên \(2\)\(⋮\)\(x-2\)
hay \(x-2\)\(\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Ta lập bảng sau:
\(x-2\) \(-2\) \(-1\) \(1\) \(2\)
\(x\) \(0\) \(1\) \(3\) \(4\)
Vậy \(x=\left\{0;1;3;4\right\}\)
Để pt có nghiệm \(\Leftrightarrow\Delta\ge0\Leftrightarrow4-4\left(m-1\right)\ge0\)\(\Leftrightarrow2\ge m\)
Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(1\right)\\x_1x_2=m-1\end{matrix}\right.\)
\(x_1^4-x_1^3=x_2^4-x_2^3\)
\(\Leftrightarrow\left(x_1^4-x_2^4\right)-\left(x_1^3-x_2^3\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)\left(x_1^2+x_2^2\right)-\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left[2\left(x_1^2+x_2^2\right)-x_1^2-x_1x_2-x_2^2\right]=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1^2+x_2^2-x_1x_2\right)=0\)
\(\Leftrightarrow x_1-x_2=0\) (2) ( vì \(x_1^2-x_1x_2+x_2^2>0;\forall x,y\))
Từ (1) (2) \(\Rightarrow x_1=x_2=1\)
\(\Rightarrow x_1x_2=m-1=1\) \(\Leftrightarrow m=2\) (Thỏa)
Vậy...
a: \(\left(-120\right):15+12\left(2x-1\right)=52\)
=>\(12\left(2x-1\right)-8=52\)
=>\(12\left(2x-1\right)=60\)
=>\(2x-1=\dfrac{60}{12}=5\)
=>2x=5+1=6
=>\(x=\dfrac{6}{2}=3\)
c: \(x+4⋮x+1\)
=>\(x+1+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
d: \(2x+7⋮x+2\)
=>\(2x+4+3⋮x+2\)
=>\(3⋮x+2\)
=>\(x+2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{-1;-3;1;-5\right\}\)
e: \(3x⋮x-1\)
=>\(3x-3+3⋮x-1\)
=>\(3⋮x-1\)
=>\(x-1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{2;0;4;-2\right\}\)
\(2x+10⋮x+1\)
\(\Rightarrow2x+2+8⋮x+1\)
\(\Rightarrow2\left(x+1\right)+8⋮x+1\)
\(\Rightarrow x+1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(x\in\left\{0;1;-2;\pm3;-5;7;-9\right\}\)
\(2x+10⋮x-1\)
\(\Rightarrow2x-2+12⋮x-1\)
\(\Rightarrow2\left(x-1\right)+12⋮x-1\)
\(\Rightarrow x-1\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
x - 1 = 1 => x = 2
x -1 = -1 => x = 0
... tg tự
Bài 1:
\(c.\) \(2x+1⋮x-1\)
\(\Leftrightarrow\left(2x-2\right)+3⋮x-1\)
\(\Leftrightarrow3⋮x-1\)
Ta có bẳng sau:
\(x-1\) | \(-1\) | \(1\) | \(3\) | \(-3\) |
\(x\) | \(0\) | \(2\) | \(4\) | \(-2\) |
14 : (2x - 1)
=> 2x+1 = { + 14, - 14 }
EZ
14 chia hết cho 2x - 1
=> 2x - 1 thuộc Ư(14)
Mà Ư(14) = {1; 2; 7; 14}
+) 2x - 1 = 1 => x = 1
+) 2x - 1 = 2 => x = 3/2
+) 2x - 1 = 7 => x = 4
+) 2x - 1 = 14 => x = 15/2
Vậy: x thuộc {1; 3/2; 4; 15/2}