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x2+3x-18=0
x2-3x+6x-18=0
x(x-3)+6.(x-3)=0
(x-3)(x+6)=0
=> x-3=0 hoặc x+6=0
•x-3=0=> x=3
•x+6=0=>x=-6
Vậy x thuộc {3;-6}
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)2x3+3x2+2x+3=0
=> (2x3+3x2)+(2x+3)=0
=> x2(2x+3)+(2x+3)=0
=> (2x+3)(x2+1)=0
=>\(\hept{\begin{cases}2x+3=0\\x^2+1=0\end{cases}}\)=>\(\hept{\begin{cases}2x=-3\\x^2=-1\end{cases}}\)=>\(\hept{\begin{cases}x=\frac{-3}{2}\\vo.nghiem\end{cases}}\)
Vậy x=-3/2
2)x2-3x-18=0
=> (x2+3x)-(6x+18)=0
=> x(x+3)-6(x+3)=0
=> (x+3)(x-6)=0
=> \(\hept{\begin{cases}x+3=0\\x-6=0\end{cases}}\)=>\(\hept{\begin{cases}x=-3\\x=6\end{cases}}\)
Vậy x=-3 hoặc x=6
3)Sai đề rồi bạn, 30 thành 30x mới đúng
x3-11x2+30x=0
=> x(x2-11x+30)=0
=> x[(x2-5x)-(6x-30)]=0
=> x[x(x-5)-6(x-5)]=0
=> x(x-5)(x-6)=0
=>\(\hept{\begin{cases}x=0\\x-5=0\\x-6=0\end{cases}}\)=>\(\hept{\begin{cases}x=0\\x=5\\x=6\end{cases}}\)
Vậy x=0 hoặc x=5 hoặc x=6
![](https://rs.olm.vn/images/avt/0.png?1311)
a:Ta có: \(x\left(x-1\right)+x=4\)
\(\Leftrightarrow x^2-x+x=4\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b: Ta có: \(3x\left(x-5\right)-2x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c: Ta có: \(5x^2-3x-2=0\)
\(\Leftrightarrow5x^2-5x+2x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: Ta có: \(x^4-11x^2+18=0\)
\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
a) x(x-1)+x=4
⇔x2=4⇔\(x=\pm2\)
b)3x(x-5)-2x+10=0
⇔3x(x-5)-2(x-5)=0
⇔(x-5)(3x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
c)5x2-3x-2=0
⇔ 5x(x-1)+2(x-1)=0
⇔ (x-1)(5x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d)x4-11x2+18=0
⇔ x2(x2-2)-9(x2-2)=0
⇔ (x2-2)(x2-9)=0
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
#)Giải :
Câu 1 :
5x(1 - 2x ) - 3x ( x+18) = 0
<=> 5x - 10x^2 - 3x^2 - 54x = 0
<=> -13x^2 - 49x = 0
<=> x= 0 hoặc x = - 49/13
Vậy x có hai giá trị là 0 và - 49/13
![](https://rs.olm.vn/images/avt/0.png?1311)
a)5x ( 1 - 2x ) - 3x ( x + 18 ) = 0
5x - 10x^2 - 3x^2 - 54x =0
-13x^2 - 49x =0
(-13x - 49)x =0
\(\Rightarrow\orbr{\begin{cases}-13x-49=0\\x=0\end{cases}\Rightarrow\orbr{\begin{cases}-13x=49\\x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{49}{13}\\x=0\end{cases}}}\)
Vậy x= -49/13 hoặc x=0
b)5x - 10x^2 - 3x^2 - 54 = 0
(câu b giống câu a)
![](https://rs.olm.vn/images/avt/0.png?1311)
5x(1-2x)-3x(x+18)=0
o) TH1 :
5x(1-2x)=0
=> x(1-2x)=0
=> x-2x2=0
=> x=0
o) TH2 :
3x(x+18)=0
=> x(x+18)=0
=> x+18=0
=> x=-18
Vậy x=0 hoặc x=-18.
![](https://rs.olm.vn/images/avt/0.png?1311)
5x(1-2x)-3x(x+18)=0
=> \(5x-10x^2-3x^2-54x=0\)
=> \(-49x-13x^2=0\)
=> \(x\left(-49-13x\right)=0\)
Suy ra
- \(x=0\)
- \(-49-13x=0\)=> \(-13x=49\)=> \(x=\frac{-49}{13}\)
Vậy \(x=\frac{-49}{13}; x=0\)
a) x2 + 3x - 18 = 0
x2 + 6x - 3x - 18 = 0
x.(x+6) - 3.(x+6) = 0
(x+6).(x-3) = 0
=> x + 6 = 0 => x = -6
x-3 =0 => x = 3
KL:./.
\(x^2+3x-18=0\)
\(\Rightarrow x^2+6x-3x-18=0\)
\(\Rightarrow x\left(x+6\right)-3\left(x+6\right)=0\)
\(\Rightarrow\left(x+6\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-6\\x=3\end{cases}}\)