\(\left\{{}\begin{matrix}ax^2+by+c=0\\cx^2+by+a=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}ax^2+by=-c\\cx^2+by=-a\end{matrix}\right.\)

vì pt có 1 nghiệm duy nhất

nên\(\dfrac{a}{c}\ne\dfrac{b}{b}\)⇔\(\dfrac{a}{c}\ne1\)⇔\(a\ne c\)

Mình nghĩ là sai đề
Cho pt \(ax^2+bx+c=0\) (1) và \(cx^2+bx+a=0\)  (2)

Lấy (1) trừ (2) ta được: \(\left(x^2-1\right)\left(a-c\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)(vì a khác c)

TH1: Giả sử nghiệm chung của hai pt là x=1

Thay x=1 vào (1) và (2) được: \(\left\{{}\begin{matrix}a+b+c=0\\a+b+c=0\end{matrix}\right.\)\(\Leftrightarrow b=-a-c\)

Áp dụng hệ thức viet vào hai pt:
\(\left\{{}\begin{matrix}x_1+1=-\dfrac{b}{a}\\x_2+1=-\dfrac{b}{c}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1=-\dfrac{b}{a}-1\\x_2=-\dfrac{b}{c}-1\end{matrix}\right.\)

Có: \(\left|x_1\right|+\left|x_2\right|>2\Leftrightarrow\left|-\dfrac{b}{a}-1\right|+\left|\dfrac{-b}{c}-1\right|>2\)

\(\Leftrightarrow\left|-\dfrac{-a-c}{a}-1\right|+\left|\dfrac{-\left(-a-c\right)}{c}-1\right|>2\)

\(\Leftrightarrow\left|\dfrac{c}{a}\right|+\left|\dfrac{a}{c}\right|>2\) \(\Leftrightarrow c^2+a^2>2\left|ac\right|\) (luôn đúng với mọi \(a\ne c\))
TH2: Giả sử x=-1 là nghiệm chung của hai pt

Thay x=-1 vào hai pt được: \(\left\{{}\begin{matrix}a-b+c=0\\c-b+a=0\end{matrix}\right.\) \(\Leftrightarrow b=a+c\)

Áp dụng viet vào hai pt có: \(\left\{{}\begin{matrix}x_1+\left(-1\right)=-\dfrac{b}{a}\\x_2+\left(-1\right)=-\dfrac{b}{c}\end{matrix}\right.\)

Khi đó: \(\left|x_1\right|+\left|x_2\right|=\left|-\dfrac{b}{a}+1\right|+\left|-\dfrac{b}{c}+1\right|\)

\(=\left|-\dfrac{a+c}{a}+1\right|+\left|-\dfrac{a+c}{c}+1\right|\)\(=\left|-\dfrac{c}{a}\right|+\left|-\dfrac{a}{c}\right|\)\(=\left|\dfrac{c}{a}\right|+\left|\dfrac{a}{c}\right|=\dfrac{c^2+a^2}{\left|ac\right|}>\dfrac{2\left|ac\right|}{\left|ac\right|}=2\)
Vậy...
 

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

\(C=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\sqrt{3^2-4\cdot\left(-7\right)}=\sqrt{37}\)

a: \(\Leftrightarrow\left(2m+4\right)^2-4m\cdot9=0\)

\(\Leftrightarrow4m^2+16m+16-36m=0\)

\(\Leftrightarrow m^2-5m+4=0\)

\(\Leftrightarrow\left(m-1\right)\left(m-4\right)=0\)

hay \(m\in\left\{1;4\right\}\)

b: \(\Leftrightarrow\left(2m-8\right)^2-4\left(m^2+m+3\right)=0\)

\(\Leftrightarrow4m^2-32m+64-4m^2-4m-12=0\)

=>-36m+52=0

=>-36m=-52

hay m=13/9

d: \(\Leftrightarrow m^2-4m\left(m+3\right)=0\)

\(\Leftrightarrow m\left(m-4m-12\right)=0\)

=>m(-3m-12)=0

=>m=0 hoặc m=-4

a) PT có nghiệm kép khi △=0

\(\Leftrightarrow\left[2\left(m+2\right)\right]^2-4.m.9=0\)

\(\Leftrightarrow4\left(m^2+4m+4\right)-36m=0\)

\(\Leftrightarrow4m^2-20m+16=0\Leftrightarrow\left[{}\begin{matrix}m=4\\m=1\end{matrix}\right.\)

Khi đó nghiệm kép của pt là \(x_1=x_2=\dfrac{-2\left(m+2\right)}{2.m}=\dfrac{-2m-4}{2m}=-1-\dfrac{2}{m}\)

+Khi m=4 thì \(x_1=x_2=-1-\dfrac{2}{4}=-\dfrac{3}{2}\)

+Khi m=1 thì \(x_1=x_2=-1-\dfrac{2}{1}=-3\)

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 \(C=\dfrac{11}{12}\)

\(3,3x^2-7x-1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=\dfrac{7}{3}\\P=x_1x_2=\dfrac{c}{a}=-\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(B=\dfrac{2x_2^2}{x_1+x_2}+2x_1\)

\(=\dfrac{2x_2^2+2x_1\left(x_1+x_2\right)}{x_1+x_2}\)

\(=\dfrac{2x_2^2+2x_1^2+2x_1x_2}{x_1+x_2}\)

\(=\dfrac{2\left(x_1^2+x_2^2\right)+2x_1x_2}{x_1+x_2}\)

\(=\dfrac{2\left(S^2-2P\right)+2P}{S}\)

\(=\dfrac{2\left(\dfrac{7}{3}^2-2\left(-\dfrac{1}{3}\right)\right)+2\left(-\dfrac{1}{3}\right)}{\dfrac{7}{3}}\)

\(=\dfrac{104}{21}\)

Vậy \(B=\dfrac{104}{21}\)

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 )