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3x2 + 2x - 1 = 0
=> 3x2 + 3x - x - 1 = 0
=> 3x(x + 1) - (x + 1) = 0
=> (3x - 1)(x + 1) = 0
=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
x2 - 5x + 6 = 0
=> x2 - 2x - 3x + 6 = 0
=> x(x - 2) - 3(x - 2) = 0
=> (x - 3)(x - 2) = 0
=> \(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
3x2 + 7x + 2 = 0
=> 3x2 + 6x + x + 2 = 0
=> 3x(x + 2) + (x + 2) = 0
=> (3x + 1)(x + 2) = 0
=> \(\orbr{\begin{cases}3x+1=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
1, \(3x^2+2x-1=0\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}}\)
2, \(x^2-5x+6=0\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
3, \(3x^2+7x+2=0\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}}\)
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a: ĐKXĐ: x<>0
\(\Leftrightarrow3x^2+10x-3x-10=0\)
=>(3x+10)(x-1)=0
=>x=-10/3 hoặc x=1
b: ĐKXĐ: \(x\in R\)
\(\Leftrightarrow4x-17=0\)
hay x=17/4
c: ĐKXĐ: \(x\ne-5\)
=>2x-5=0
hay x=5/2
d: ĐKXĐ: x<>-2/3
\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=5\)
\(\Leftrightarrow6x^2+4x-3x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
=>(6x+7)(x-1)=0
=>x=1 hoặc x=-7/6
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+1=0\\2x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-1\\x=-\frac{3}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\x^2=-1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(S=\left\{\dfrac{3}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-1\right)^2+\left(2x-x\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1+2x-x\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(4x-x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\\3x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
3x2+ 2x- 1= 0
<=> 3x2+ 3x- x- 1= 0
<=> (3x2+ 3x)- (x+ 1)= 0
<=> 3x(x+ 1)- (x+ 1)= 0
<=> (x+1)(3x-1)= 0
<=>\(\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\\3x=1\end{cases}}\)<=> \(\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)
Vậy pt có 2 nghiệm S={ -1; 1/3}
\(3x^2+2x-1=0\)
\(\Rightarrow3x^2+3x-x-1=0\)
\(\Rightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}3x-1=0\\x+1=0\end{cases}\hept{\begin{cases}3x=1\Rightarrow x=\frac{1}{3}\\x=1\end{cases}}}\)
Vậy x=1/3, x=1