X2 - X + 2 (X - 2) =0
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a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
a) (x - 3)2 - 5.(x - 2) + 5 = 0.
<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0
<=> x^2 - 11x + 24 = 0
<=> (x-3)(x-8)=0
<=> x = 3 hoặc x = 8
a: (x-3)(x-2)<0
=>x-2>0 và x-3<0
=>2<x<3
b: \(\left(x+3\right)\left(x+4\right)\left(x^2+2\right)\ge0\)
=>(x+3)(x+4)>=0
=>x+3>=0 hoặc x+4<=0
=>x>=-3 hoặc x<=-4
c: \(\dfrac{x-1}{x-2}\ge0\)
=>x-2>0 hoặc x-1<=0
=>x>2 hoặc x<=1
d: \(\dfrac{x+3}{2-x}>=0\)
=>\(\dfrac{x+3}{x-2}< =0\)
=>x+3>=0 và x-2<0
=>-3<=x<2
e, \(\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2=0\Leftrightarrow\left(x-2\right)\left(x+2+x-2\right)=0\Leftrightarrow x=0;x=2\)
f, \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x-1\right)^2=0\Leftrightarrow x=1;x=-1\)
g, \(x^2\left(x-3\right)+4\left(3-x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\Leftrightarrow x=2;x=-2;x=3\)
h, \(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\Leftrightarrow x=4;x=-\dfrac{2}{3}\)
a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
x-2 | 1 | -1 | 3 | -3 |
x | 3 | 1 | 5 | -1 |
b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
x-2 | 1 | -1 | 13 | -13 |
x | 3 | 1 | 15 | -11 |
c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+7 | 1 | -1 | 2 | -2 |
x | -6 | -8 | -5 | -9 |
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
X2 - X + 2 (X - 2) =0
<=> X2 -X+2X-4=0
<=> X2+X-4=0
<=> x= \(\dfrac{-1\pm\sqrt{17}}{2}\)
x2 - x + 2 ( x - 2 ) = 0
x ( x - 1 ) + 2 ( x - 1 ) - 2 = 0
( x - 1 )( x + 2 ) - 2 = 0
( x - 1 )( x + 2 ) = 2
Ta có bảng sau:
Vậy x = 0 thỏa mãn đề bài