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a,x(x-2)+x-2=0
⇔ (x-2)(x+1)=0
⇔ x=2;x=-1
b,x3+x2+x+1=0
⇔ x2(x+1)+x+1=0
⇔ (x+1)(x2+1)=0
⇔ x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x-1\right)=0\)
=>x=-1 hoặc x=1
b: \(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
hay \(x\in\left\{-1;2;-2\right\}\)
c: \(x^3+x^2+4=0\)
\(\Leftrightarrow x^3+2x^2-x^2-2x+2x+4=0\)
\(\Leftrightarrow\left(x+2\right)\cdot\left(x^2-x+2\right)=0\)
=>x+2=0
hay x=-2
e: \(\Leftrightarrow x^4-2x^3-3x^3+6x^2-x^2+2x+3x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-3x^2-x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)=0\)
hay \(x\in\left\{2;3;-1;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow\left(x+3\right)\left(x+3-x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow x=0\left(x^2+4>0\right)\)
\(a,x^2+2.x.3+3^2-\left(x^2-3^2\right)=0\)
\(x^2+6x+9-x^2+9=0\)
\(6x+18=0\)
\(6x=-18\)
\(x=-3\)
Vậy x=-3
\(b,5x^3+20x=0\)
\(5x\left(x^2+4\right)=0\)
\(Th1:5x=0=>x=0\)
\(Th2:x^2+4=0\)
\(x^2=-4\)(vô lý)
Vậy x=0
giải các Phương trình sau
a) (5x+3)(x2+1)(x-1)=0
b) (4x-1)(x-3)-(x-3)(5x+2)=0
c) (x+6)(3x-1)+x2-36 =0
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>(5x+3)(x-1)=0
=>x=1 hoặc x=-3/5
b: =>(x-3)(4x-1-5x-2)=0
=>(x-3)(-x-3)=0
=>x=-3 hoặc x=3
c: =>(x+6)(3x-1+x-6)=0
=>(x+6)(4x-7)=0
=>x=7/4 hoặc x=-6
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta có:
\(x^2-7x+12=0\)
\(\Leftrightarrow x^2-3x-4x+12=0\)
\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
b) Ta có:
\(x\left(x-4\right)-3\left(4-x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 10:
a) (x+2)2 -x(x+3) + 5x = -20
=> x2 + 4x + 4 - x2 - 3x + 5x = -20
=> 6x = -20 + (-4)
=> 6x = -24
=> x = -4
b) 5x3-10x2+5x=0
=>5x(x2-2x+1)=0
=>5x(x-1)2 =0
=> 5x=0 hoặc (x-1)2=0
=>x=0 hoặc x=1
c) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
=> (x2 - 1)[(x2 - 1)2 - (x4 + x2 + 1)] = 0
<=> (x2 - 1)(x4 - 2x2 + 1 - x4 - x2 - 1) = 0
<=> (x2 - 1)(-3x2) = 0
<=> (x2 - 1)=0 hoặc (-3x2) =0
<=> x2=1 hoặc x2=0
<=> x=−1;1 hoặc x=0
d)
(x+1)3−(x−1)3−6(x−1)2=-19
⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0
⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0
⇔12x+13=0⇔12x+13=0
⇔12x=-13
⇔x=-23/12
Học tốt nhé:333
![](https://rs.olm.vn/images/avt/0.png?1311)
a
\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)
b
\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)
c
\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)
a: =>(2x+15)(x^2+4)=0
=>2x+15=0
=>2x=-15
=>x=-15/2
b; =>(x-2)(5x-3)=0
=>x=2 hoặc x=3/5
c: =>(x+3)(2-x)=0
=>x=2 hoặc x=-3
\(5x\left(x-3\right)-x+3=0\)
<=> \(\left(x-3\right)\left(5x-1\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)
Vay..........