Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a.Bạn thế vào nhé
b.\(\Delta=3^2-4m=9-4m\)
Để pt vô nghiệm thì \(\Delta< 0\)
\(\Leftrightarrow9-4m< 0\Leftrightarrow m>\dfrac{9}{4}\)
c.Ta có: \(x_1=-1\)
\(\Rightarrow x_2=-\dfrac{c}{a}=-m\)
d.Theo hệ thức Vi-ét, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-3\\x_1.x_2=m\end{matrix}\right.\)
1/ \(x_1^2+x_2^2=34\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=34\)
\(\Leftrightarrow\left(-3\right)^2-2m=34\)
\(\Leftrightarrow m=-12,5\)
..... ( Các bài kia tương tự bạn nhé )
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1x_2=-8\end{matrix}\right.\)
\(M=x_1\left(1-x_2\right)+x_2\left(1-x_1\right)\)
\(=x_1+x_2-2x_1x_2\)
\(=-2-2.\left(-8\right)=14\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. Em tự giải
b. Pt có 2 nghiệm khi \(\Delta=9-4\left(m-4\right)\ge0\Rightarrow m\le\dfrac{25}{4}\)
Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-3\\x_1x_2=m-4\end{matrix}\right.\)
c.
\(x_1^3+x_2^3=8\)
\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=8\)
\(\Leftrightarrow\left(-3\right)^3-3.\left(-3\right).\left(m-4\right)=8\)
\(\Leftrightarrow m=\dfrac{71}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{4}{3}\\x_1.x_2=\dfrac{1}{3}\end{matrix}\right.\)
\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_1-1\right)\left(x_2-1\right)}\)
\(=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_1-x_2+1}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}=\dfrac{\dfrac{22}{9}}{\dfrac{8}{3}}=\dfrac{11}{12}\)
\(1,3x^2+4x+1=0\)
Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :
\(\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}=-\dfrac{4}{3}\\P=x_1x_2=\dfrac{c}{a}=\dfrac{1}{3}\end{matrix}\right.\)
Ta có :
\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}\)
\(=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_2-1\right)\left(x_1-1\right)}\)
\(=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_2-x_1+1}\)
\(=\dfrac{\left(x_1^2+x_2^2\right)-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{S^2-2P-S}{P-S+1}\)
\(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}\)
\(=\dfrac{11}{12}\)
Vậy \(C=\dfrac{11}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x1+x2=3; x1*x2=-7
B=(x1+x2)^2-2x1x2
=9-2*(-7)=23
D=(x1+x2)^3-3x1x2(x1+x2)
=3^3-3*(-7)*3
=27+63=90
F=9x1x2+3(x1^2+x2^2)+x1x2
=10x1x2+3*23
=10*(-7)+69
=-1
\(C=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\sqrt{3^2-4\cdot\left(-7\right)}=\sqrt{37}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1a. Bạn tự giải
b/ \(\Delta=9-4\left(4m-1\right)=13-16m\)
Để pt có 2 nghiệm
\(\Leftrightarrow13-16m\ge0\Rightarrow m\le\frac{13}{16}\)
2.
\(\Delta'=\left(m+7\right)^2-\left(m^2-4\right)=14m+53\)
Để pt có 2 nghiệm \(\Rightarrow14m+53\ge0\Rightarrow m\ge-\frac{53}{14}\)
Theo Viet ta có: \(x_1+x_2=2\left(m+7\right)\)
\(\Rightarrow2\left(m+7\right)=10\Rightarrow m+7=5\Rightarrow m=-2\) (thỏa mãn)