2x2+ 5x – 3 = 0
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\(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy \(S=\left\{-1;-\dfrac{3}{2}\right\}\)
\(\left(x-\sqrt{2}\right)-3\left(x^2-2\right)=0\)
\(\Leftrightarrow x-\sqrt{2}-3x^2+6=0\)
\(\Leftrightarrow-3x^2+x+6-\sqrt{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{1}{6}-\dfrac{\sqrt{73-3\sqrt{32}}}{6}\\x_2=\dfrac{\sqrt{73-3\sqrt{32}}}{6}+\dfrac{1}{6}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)
\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\left(5x+3\right).5\left(3x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chọn C
Ta có f(x) + g(x) = (2x2 - 5x - 3) + (-2x2 - 2x + 1) = -7x - 2
Cho -7x - 2 = 0 ⇒ x = -2/7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2x^2+x=0\)
\(x\left(2x+1\right)=0\)
\(\left[{}\begin{matrix}x=0\\2x=-1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-\frac{1}{2}\end{matrix}\right.\)
\(b,-0,4x^2+1,2x=0\)
\(x\left[\left(0,4x\right)-\left(1,2\right)\right]=0\)
\(\left[{}\begin{matrix}x=0\\0,4x-1,2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\0,4x=1,2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=\frac{3}{10}\end{matrix}\right.\)
\(c,7x^2-5x=0\)
\(x\left(7x-5\right)=0\)
\(\left[{}\begin{matrix}x=0\\7x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=\frac{5}{7}\end{matrix}\right.\)
\(e,-2x^2-11x=0\)
\(x\left(2x+11\right)=0\)
\(\left[{}\begin{matrix}x=0\\2x+11=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=\frac{11}{2}\end{matrix}\right.\)
ko có bạn thì mình chết từ lâu rồi cảm ơn bạn nhiều
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
a) \(2x^4-4x^3+2x^2\)
\(=2x^2\left(x^2-2x+1\right)\)
\(=2x^2\left(x-1\right)^2\)
b) \(2x^2-2xy+5x-5y\)
\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)
\(=2x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(2x+5\right)\)
2 .
a,
\(4x\left(x-3\right)-x+3=0\)
⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)
⇒\(\left(x-3\right)\left(4x-1\right)=0\)
⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)
b,
\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)
⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0
⇒\(\left(x-4\right)\left(3x-2\right)=0\)
⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)
ta có \(\Delta\)= 52-4.(-3).2 =49>0, \(\sqrt{\Delta}\)=7
Vậy phương trình có hai nghiệm phân biệt : x1=\(\frac{-5-7}{2.2}\)=-3; x2=\(\frac{-5+7}{2.2}\)=0,5