giải pt -2x2+3x-1
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-3x+3+x\right)-2x^2=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)^2+x\left(x^2-3x+3\right)-2x^2=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)^2-x\left(x^2-3x+3\right)+2x\left(x^2-3x+3\right)-2x^2=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-3x+3-x\right)+2x\left(x^2-3x+3-x\right)=0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2-x+3=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>7-x=0
hay x=7
b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>-x+7=0
hay x=7
b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Delta=2^2-4\cdot1\cdot\left(-30\right)=124\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-2-2\sqrt{31}}{2}=-1-\sqrt{31}\\x_2=-1+\sqrt{31}\end{matrix}\right.\)
b: \(2x^2-3x-5=0\)
\(\Leftrightarrow2x^2-5x+2x-5=0\)
=>(2x-5)(x+1)=0
=>x=5/2 hoặc x=-1
a.\(x^2+2x-30=0\)
\(\Delta=2^2-4.\left(-30\right)=4+120=124>0\)
=> pt có 2 nghiệm
\(\left\{{}\begin{matrix}x=\dfrac{-2+\sqrt{124}}{2}=\dfrac{-2+2\sqrt{31}}{2}=-1+\sqrt{31}\\x=\dfrac{-2-\sqrt{124}}{2}=-1-\sqrt{31}\end{matrix}\right.\)
b.\(2x^2-3x-5=0\)
Ta có: a-b+c=0
\(\Rightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{5}{2}\end{matrix}\right.\)( vi-ét )
![](https://rs.olm.vn/images/avt/0.png?1311)
PT có 2 no âm phân biệt \(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(-3\right)^2-4\left(-2\right)\left(-m+1\right)>0\\x_1+x_2=\dfrac{3}{-2}< 0\\x_1x_2=\dfrac{-m+1}{-2}>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}17-8m>0\\-m+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{17}{8}\\m>1\end{matrix}\right.\Leftrightarrow1< m< \dfrac{17}{8}\)
Mình chưa hiểu ngay chỗ \(\dfrac{-m+1}{-2}\)> 0 ➜ -m+1<0 v á.
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{3}{2}\\x_1x_2=-\dfrac{1}{2}\end{matrix}\right.\)
\(B=\dfrac{4x_1-1}{x_2}+\dfrac{4x_2-1}{x_1}=\dfrac{4x_1^2-x_1+4x_2^2-x_2}{x_1x_2}\)
\(=\dfrac{4\left(x_1+x_2\right)^2-8x_1x_2-\left(x_1+x_2\right)}{x_1x_2}=\dfrac{4.\left(-\dfrac{3}{2}\right)^2-8.\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{2}\right)}{-\dfrac{1}{2}}=-29\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(3x-4\right)=2x^2+1\)
\(\Leftrightarrow3x^2-4x-2x^2-1=0\)
\(\Leftrightarrow x^2-4x-1=0\)
Theo Vi - ét, ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=4\\x_1x_2=\dfrac{c}{a}=-1\end{matrix}\right.\)
Ta có :
\(A=x_1^2+x_2^2+3x_1x_2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2+3x_1x_2\)
\(=\left(x_1+x_2\right)^2+x_1x_2\)
\(=4^2-1\)
\(=16-1\)
\(=15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-2x^2 + 3x -1 = -2x^2 +2x + x -1
=-2x(x-1) + (x - 1)
=(-2x+1)(x-1)
Do không có vế thứ 2 nên mình chỉ giải đc đến đây