Giải phương trình sau : \(3x^2+2x−1=0\).
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\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
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\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x-1}=-x^2+3x-1\\ \Leftrightarrow2x-1=\left(-x^2+3x-1\right)^2=\left(x^2-3x+1\right)^2\\ \Leftrightarrow2x-1=x^4+9x^2+1-6x^3-6x+2x^2\\ \Leftrightarrow x^4-6x^3+11x^2-8x+2=0\\ \Leftrightarrow x^4-x^3-5x^3+5x^2+6x^2-6x-2x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x^3-5x^2+6x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3-x^2-4x^2+4x+2x-2\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x^2-4x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x^2-4x+2=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=16-8=8\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4-2\sqrt{2}}{2}=2-\sqrt{2}\left(tm\right)\\x=\dfrac{4+2\sqrt{2}}{2}=2+\sqrt{2}\left(tm\right)\end{matrix}\right.\\ S=\left\{1;2-\sqrt{2};2+\sqrt{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`3x+7=0`
`<=>3x=-7`
`<=>x=-7/3`
Vậy `S={-7/3}`
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`2x(x-2)+2x(5-3x)=0`
`<=>2x(x-2+5-3x)=0`
`<=>2x(3-2x)=0`
`@TH1:2x=0<=>x=0`
`@TH2: 3-2x=0<=>2x=3<=>x=3/2`
Vậy `S={0;3/2}`
3x+7=0
\(\Leftrightarrow3x=-7\Leftrightarrow x=-\dfrac{7}{3}\)
2x(x-2)+2x(5-3x)=0
\(\Leftrightarrow2x\left(x-2+5-3x\right)=0\)
\(\Leftrightarrow2x\left(-2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-3}{-2}=\dfrac{3}{2}\end{matrix}\right.\)
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a) \(4x-9=0\) \(\Leftrightarrow4x=9\) \(\Leftrightarrow x=\dfrac{9}{4}\)
Vậy \(x=\dfrac{9}{4}\)
b) \(-2x+50=0\) \(\Leftrightarrow2x=50\) \(\Leftrightarrow x=25\)
Vậy \(x=25\)
c) \(3x+11=0\) \(\Leftrightarrow3x=-11\) \(\Leftrightarrow x=-\dfrac{11}{3}\)
Vậy \(x=-\dfrac{11}{3}\)
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1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ ( x-1) (2x+1) =0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-0,5\end{matrix}\right.\)
2/ x (2x-1) (3x+15) =0
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-5\end{matrix}\right.\)
3/ (2x-6) (3x+4).x=0
\(\Rightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)
4/ (2x-10)(x2+1)=0
\(\Rightarrow\left[{}\begin{matrix}2x-10=0\\x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x^2=-1\left(loại\right)\end{matrix}\right.\)
5/ (x2+3) (2x-1) =0
\(\Rightarrow\left[{}\begin{matrix}x^2+3=0\\2x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x^2=-3\left(loại\right)\\x=0,5\end{matrix}\right.\)
6/ (3x-1) (2x2 +1)=0
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\2x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x^2=-0,5\left(loại\right)\end{matrix}\right.\)
1: Ta có: \(\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
2: Ta có: \(x\left(2x-1\right)\left(3x+15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-5\end{matrix}\right.\)
3: Ta có: \(\left(2x-6\right)\left(3x+4\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(x^3-3x^2+2x=0\)
\(\Leftrightarrow x\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
Vậy: S={0;1;2}
2) Ta có: \(\dfrac{x^2-x-1}{x+1}=2x-1\)
\(\Leftrightarrow x^2-x-1=\left(2x-1\right)\left(x+1\right)\)
\(\Leftrightarrow x^2-x-1=2x^2+2x-x-1\)
\(\Leftrightarrow x^2-x-1-2x^2-x+1=0\)
\(\Leftrightarrow-x^2-2x=0\)
\(\Leftrightarrow-x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy: S={0;-2}
3x2+2x=0
<=>x(3x+2)=0
<=>x=0 hoặc 3x+2=0
từ đó bạn giải ra x thuộc{0;-2/3}
chúc bạn học tốt và nhớ tích đúng cho mình
![](https://rs.olm.vn/images/avt/0.png?1311)
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a/ 4x + 20 = 0
⇔4x = -20
⇔x = -5
Vậy phương trình có tập nghiệm S = {-5}
b/ 2x – 3 = 3(x – 1) + x + 2
⇔ 2x-3 = 3x -3+x+2
⇔2x – 3x = -3+2+3
⇔-2x = 2
⇔x = -1
Vậy phương trình có tập nghiệm S = {-1}
câu tiếp theo
a/ (3x – 2)(4x + 5) = 0
3x – 2 = 0 hoặc 4x + 5 = 0
- 3x – 2 = 0 => x = 3/2
- 4x + 5 = 0 => x = – 5/4
Vậy phương trình có tập nghiệm S= {-5/4,3/2}
b/ 2x(x – 3) – 5(x – 3) = 0
=> (x – 3)(2x -5) = 0
=> x – 3 = 0 hoặc 2x – 5 = 0
* x – 3 = 0 => x = 3
* 2x – 5 = 0 => x = 5/2
Vậy phương trình có tập nghiệm S = {0, 5/2}
3x2 + 2x - 1 = 0
<=> 3x2 + 3x - x - 1 = 0
<=> 3x(x + 1) - (x + 1) = 0
<=> (3x - 1)(x + 1) = 0
<=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\) Vậy S = {-1; 1/3}
Trả lời:
3x^2 +2x -1 =0
<=> 3x^2 +3x -x -1 =0
<=> ( 3x^2 +3x ) -( x +1 )=0
<=> 3x (x +1 ) - ( x +1 ) =0
<=>( 3x -1 ) ( x +1 ) =0
Đến đây bạn tự làm tiếp nhé
#Học tốt:))