Ôn tập chương III

a: \(9m+9+3m-3+m=0\)

=>13m=-6

hay m=-6/13

\(a,\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\\dfrac{c}{a}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}25-12\left(2m+1\right)\ge0\\\dfrac{2m+1}{3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le\dfrac{13}{24}\\m< -\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow m< -\dfrac{1}{2}\)

\(b,\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\dfrac{c}{a}>0\\-\dfrac{b}{a}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{13}{24}\\m>-\dfrac{1}{2}\\-\dfrac{5}{3}< 0\left(đúng\right)\end{matrix}\right.\Leftrightarrow-\dfrac{1}{2}< m< \dfrac{13}{24}\)

\(c,\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\dfrac{c}{a}>0\\-\dfrac{b}{a}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{13}{24}\\m>-\dfrac{1}{2}\\-\dfrac{5}{3}>0\left(\text{vô lí}\right)\end{matrix}\right.\Leftrightarrow m\in\varnothing\)

\(d,\Leftrightarrow\Delta\ge0\Leftrightarrow m\le\dfrac{13}{24}\\ \text{Viét: }\left\{{}\begin{matrix}x_1+x_2=-\dfrac{5}{3}\\x_1x_2=\dfrac{2m+1}{3}\end{matrix}\right.\\ x_1^3+x_2^3+10\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=10\\ \Leftrightarrow-\dfrac{125}{27}-\left(2m+1\right)\left(-\dfrac{5}{3}\right)=10\\ \Leftrightarrow\dfrac{5}{3}\left(2m+1\right)=\dfrac{395}{27}\\ \Leftrightarrow2m+1=\dfrac{79}{9}\Leftrightarrow m=\dfrac{35}{9}\left(ktm\right)\\ \Leftrightarrow m\in\varnothing\)

Vậy ko có m thỏa đề

a: Để phương trình có hai nghiệm trái dấu thì 2m+1<0

hay m<-1/2

a: Để pt có hai nghiệm trái dấu thì 2m+1<0

hay m<-1/2

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+5x+3-x^2+2x-1=0\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+7x+2=0\\x>=1\end{matrix}\right.\)

\(\Leftrightarrow x\in\varnothing\)

2x\(^2\)+5x+3=(x-1)\(^2\)⇔2x\(^2\)+5x+3=x\(^2\)-2x+1⇔x\(^2\)+7x+2=0⇔

x=∅

\(\Leftrightarrow x^4+x-4x-4=0\)

=>x+1=0

hay x=-1

\(x^4-3x-4=0\\ x^4+x-4x-4=0\\ x^3\left(x+1\right)-4\left(x+1\right)=0\\ \left(x^3-4\right)\left(x+1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x^3-4=0\Leftrightarrow x^3=4\Leftrightarrow x=\sqrt[3]{4}\\x+1=0\Leftrightarrow x=-1\end{matrix}\right.\)

Dạ em lớp 8 sao giải được bài 11 vậy ạ (hay đây là bài lớp 8)

Câu 5: B

Câu 6: A

Câu 4: A

Câu 3: A

Chọn B