Bài 6: Hệ thức Vi-et và ứng dụng

dễ thấy \(\Delta=m^2-2m+1=\left(m-1\right)^2\ge0\)

\(x_1=\dfrac{3m-1+\left|m-1\right|}{2}\) ;\(x_2=\dfrac{3m-1-\left|m-1\right|}{2}\)

xét m\(\ge1\)

\(\Rightarrow x_1=2m-1;x_2=m\)

\(x_1=x_2^2\Rightarrow m=1\left(tm\right)\)

xét m\(\le1\)

\(\Rightarrow x_2=2m-1;x_1=m\)

\(x_1=x_2^2\Rightarrow\left[{}\begin{matrix}m=1\\m=\dfrac{1}{4}\end{matrix}\right.\)

\(\Delta'=2^2-1\cdot\left(m+1\right)\)

\(=4-m-1\\ =3-m\)

Để pt có nghiệm \(\Leftrightarrow\Delta'\ge0\Leftrightarrow3-m\ge0\Leftrightarrow m\le3\)

Với \(m\le3\) theo vi-ét ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-4\\x_1\cdot x_2=m+1\end{matrix}\right.\)

Ta có : \(x^2_1+x^2_2=10\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\\ \Leftrightarrow\left(-4\right)^2-2\left(m+1\right)=10\)

\(\Leftrightarrow16-2m-2-10=0\\ \Leftrightarrow4-2m=0\)

\(\Leftrightarrow m=2\) (TM \(m\le3\))

Vậy........................

\(\Delta=\left(2m-1\right)^2-4\cdot2\cdot\left(m-1\right)\)

\(=4m^2-4m+1-8m+8\\ =4m^2-12m+9\\ =\left(2m-3\right)^2\ge0\forall x\)

Vì pt luôn có nghiệm với mọi x , theo vi-ét ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-2m+1\\x_1\cdot x_2=m-1\end{matrix}\right.\)

Ta có : \(4x_1^2+2x_1x_2+4x_2^2=1\)

\(\Leftrightarrow\left(4x_1^2+8x_1x_2+4x_2^2=1\right)-6x_1x_2=1\\ \Leftrightarrow4\left(x_1+x_2\right)^2-6x_1x_2=1\)

\(\Leftrightarrow4\left(-2m+1\right)^2-6\cdot\left(m-1\right)=1\\ \Leftrightarrow4\left(4m^2-4m+1\right)-6m+6=1\)

\(\Leftrightarrow16m^2-16m+4-6m+6-1=0\\ \Leftrightarrow16m^2-24m+5=0\)

\(\Delta'_m=\left(-12\right)^2-16\cdot5=144-80=64\)

\(\Rightarrow\sqrt{\Delta'_m}=8\)

Vì \(\Delta'>0\) nên pt có 2 nghiệm phân biệt

\(\Rightarrow x_1=\dfrac{12+8}{16}=\dfrac{5}{4}\)

\(x_2=\dfrac{12-8}{16}=\dfrac{1}{4}\)

Vậy..............................

Ta có: \(\Delta=\left(2m-1\right)^2-4.2.\left(m-1\right)=4m^2-4m+1-8m+8=4m^2-12m+9=\left(2m-3\right)^2\ge0\)

Vậy phương trình luôn có 2 nghiệm.

Theo định lí Vi-et, ta có:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2m-1}{-2}=\dfrac{1-2m}{2}\\x_1.x_2=\dfrac{m-1}{2}\end{matrix}\right.\)

\(\Rightarrow x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\left(\dfrac{1-2m}{2}\right)^2-2.\dfrac{m-1}{2}=\dfrac{4m^2-4m+1}{4}-\left(m-1\right)=\dfrac{4m^2-4m+1-4m+4}{4}=\dfrac{4m^2-8m+5}{4}\)

\(\Rightarrow4x_1^2+2x_1x_2+4x_2^2=4.\dfrac{4m^2-8m+5}{4}+2.\dfrac{m-1}{2}=4m^2-8m+5+m-1=4m^2-7m+4\)

Để \(4x_1^2+2x_1x_2+4x_2^2=1\)thì\(4m^2-7m+4=1\)

\(\Leftrightarrow4m^2-7m+3=0\)

Ta có: 4-7+3=0

=> Phương trình có 2 nghiệm

\(m_1=1;m_2=\dfrac{3}{4}\)

Vậy với m=1 hoặc m=\(\dfrac{3}{4}\) thì phương trình có 2 nghiệm x1;x2 thỏa mãn\(4x_1^2+2x_1x_2+4x_2^2=1\)

Đúng thì tick nhé

PT 2x²+(2m-1)x+m-1=0

\(\Delta\)=(2m-1)²-4.2.(m-1)=4m²-4m+1-8m+8=4m²-12m+9=(2m-3)²

vì (2m-3)²\(\ge\)0 với mọi m=>\(\Delta\ge0\) nên PT có 2 nghiệm phân biệt

theo định lí Vi-ét có \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-2m+1}{2}\\x_1x_2=\dfrac{m-1}{2}\end{matrix}\right.\)

ta có 4x₁²+2x₁x₂+4x₂²=1

\(\Leftrightarrow4\left(x_1^2+x_2^2\right)+2x_1x_2=1\)

\(\Leftrightarrow4\left(x_1^2+2x_1x_2+x_2^2-2x_1x_2\right)+2x_1x_2=1\)

\(\Leftrightarrow4\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+2x_1x_2=1\)

\(\Rightarrow4\left[\left(\dfrac{-2m+1}{2}\right)^2-2.\dfrac{m-1}{2}\right]+2.\dfrac{m-1}{2}=1\)

\(\Leftrightarrow\left(-2m+1\right)^2-4\left(m-1\right)+m-1-1=0\)

\(\Leftrightarrow4m^2-4m+1-4m+4+m-2=0\)

\(\Leftrightarrow4m^2-7m+3=0\)

có a+b+c=4-7+3=0

\(\left\{{}\begin{matrix}m_1=1\\m_2=\dfrac{3}{4}\end{matrix}\right.\)(thỏa mãn)

vậy m=1;m=\(\dfrac{3}{4}\) thỏa mãn yêu cầu đề bài

-theo hệ thức vi-et ta có:

x₁+x₂=-b/a=3-m (1)

x₁x₂=c/a=1-m (2)

-ta lại có:

\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=1\Leftrightarrow\dfrac{x_1^2+x_2^2}{\left(x_1x_2\right)^2_{ }_{ }}=\dfrac{\left(x_1x_2\right)^2}{\left(x_1x_2\right)^2}\Leftrightarrow x_1^2+x_2^2=\left(x_1x_2\right)^2\Leftrightarrow x_1^2+2x_1x_2+x_2^2-2x_1x_2=\left(x_1x_2\right)^2\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=\left(x_1x_2\right)^2\left(3\right)\)

thế (1), (2) vào (3):

(3-m)²-2(1-m)=(1-m)²

giải hệ ta đc:

m=3

Mình làm riết nhưng không chắc chắn kết quả, có khi lại sai. Mong mấy bạn giúp đỡ. Mình xin cảm ơn ^^

D

Lời giải

Với m =1 hệ vô nghiêm

với m khác 1

\(\left(m-1\right)^2\left[\left(m+2\right)^2-4m\right]\ge0\)

\(\Rightarrow\left(m-1\right)^4\ge0\) => đk m khác 1

với m khác 1

ta có

\(\left\{{}\begin{matrix}x_1.x_2=\dfrac{m}{\left(m-1\right)^2}\\x_1+x_2=\dfrac{\left(m+2\right)\left(m-1\right)}{\left(m-1\right)^2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x_1.x_2=\dfrac{1}{m-1}+\dfrac{1}{\left(m-1\right)^2}\\\left(x_1+x_2\right)-1=\dfrac{3}{\left(m-1\right)}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x_1.x_2=\dfrac{1}{m-1}+\dfrac{1}{\left(m-1\right)^2}\\\dfrac{x_1+x_2-1}{3}=\dfrac{1}{\left(m-1\right)}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x_1.x_2-\left(\dfrac{x_1+x_2-1}{3}\right)=\dfrac{1}{\left(m-1\right)^2}\\\left(\dfrac{x_1+x_2-1}{3}\right)^2=\dfrac{1}{\left(m-1\right)^2}\end{matrix}\right.\)

\(\left(\dfrac{x_1+x_2-1}{3}\right)^2+\left(\dfrac{x_1+x_2-1}{3}\right)-x_1x_2=0\)

\(\Delta=\left[-\left(\cdot2m+1\right)\right]^2-4\cdot1\cdot\left(m^2+3\right)\)

\(=4m^2+4m+1-4m^2-12\)

\(=4m-11\)

Để pt có nghiệm \(\Leftrightarrow\Delta\ge0\Leftrightarrow4m-11\ge0\)

\(\Leftrightarrow4m\ge11\\ \Leftrightarrow m\ge\dfrac{11}{4}\)

Với \(m\ge\dfrac{11}{4}\) theo vi-ét ta có :

\(\left\{{}\begin{matrix}x_1+x_2=2m+1\\x_1\cdot x_2=m^2+3\end{matrix}\right.\)

Ta có : \(x_1^2+x_2^2=1\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=1\)

\(\Leftrightarrow\left(2m+1\right)^2-2\left(m^2+3\right)=1\\ \Leftrightarrow4m^2+4m+1-2m^2-6=1\)

\(\Leftrightarrow2m^2+4m-6=0\)

Nhận xét : \(a+b+c=2+4-6=0\)

\(\Rightarrow m_1=1\) (KTM \(m\ge\dfrac{11}{4}\) )

\(m_2=\dfrac{-6}{2}=-3\) (KTM \(m\ge\dfrac{11}{4}\) )

Vậy ko có giá trị nào của m để \(x^2_1+x_2^2=1\)

Mình nhầm tí:

\(x_1^2+x_2^2=6\)

HD

m thỏa mãn hệ

\(\left\{{}\begin{matrix}x_1^2+x^2_2=\left(x_1+x_2\right)^2-2x_1x_2=6\\x_1+x_2=2m\\x_1.x_2=m^2-m-3\\m^2+m+3\ge0\end{matrix}\right.\)

Giải hệ => m

\(x^2+2\left(m+1\right)x+2m-4=0\)

a) \(\Delta^'=b'^2-ac=\left(m+1\right)^2-1.\left(2m-4\right)\)

=\(m^2+2m+1-2m+4\)

\(=m^2+5\)

pt có 2 nghiệm phân biệt khi \(\Delta'>0\)

Ta có: \(m^2\ge0\)

\(\Leftrightarrow m^2+5>0\)

Do đó pt có 2 nghiệm phân biệt

b) Theo định lý vi ét:

\(x_1+x_2=-\dfrac{b}{a}=\dfrac{-2\left(m+1\right)}{1}=-2m-2\)

\(x_1.x_2=\dfrac{c}{a}=\dfrac{2m-4}{1}=2m-4\)

Mà \(x_1^2+x_2^2=12\)

\(\Rightarrow\left(x_1+x_2\right)^2-2x_1x_2=12\)

=>\(\left(-2m-2\right)^2-2\left(2m-4\right)=12\)

\(\Rightarrow\left(-2m\right)^2-2.2m.2+2^2-4m+8=12\)

\(\Rightarrow4m^2-8m+4-4m+8=12\)

\(\Rightarrow4m^2-12m+12=12\)

\(\Rightarrow4m^2-12m+12-12=0\)

\(\Rightarrow4m^2-12m=0\)

=>\(2m.\left(2m-6\right)=0\)

=>\(\left[{}\begin{matrix}2m=0\\m-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=3\end{matrix}\right.\)

vậy với m=0, m=3 thì \(x_1^2+x^2_2=12\)

\(\Delta'=\left[-2\left(m+2\right)\right]^2-2\cdot\left(2m^2+1\right)\)

\(=4\left(m^2+4m+4\right)-4m^2-2\\ =4m^2+16m+16-4m^2-2\\ =16m+14\)

Để pt có nghiệm \(\Leftrightarrow\Delta'\ge0\Leftrightarrow16m+14\ge0\)

\(\Leftrightarrow16m\ge-14\\ m\ge\dfrac{-7}{8}\)

Với \(m\ge\dfrac{-7}{8}\) theo vi-ét ta có :

\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+2\right)\\x_1x_2=\dfrac{2m^2+1}{2}\end{matrix}\right.\)

Ta có : \(x^2_1+x_2^2=\dfrac{15}{2}\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=\dfrac{15}{2}\)

\(\Leftrightarrow\left[2\left(m+2\right)\right]^2-2\cdot\dfrac{2m^2+1}{2}=\dfrac{15}{2}\)

\(\Leftrightarrow4\left(m^2+4m+4\right)-2m^2-1=\dfrac{15}{2}\)

\(\Leftrightarrow4m^2+16m+16-2m^2-1-\dfrac{15}{2}=0\)

\(\Leftrightarrow2m^2+16m+\dfrac{15}{2}=0\)

\(\Leftrightarrow4m^2+32m+15=0\)

\(\Delta_m=32^2-4\cdot4\cdot15=1024-240=784>0\)

\(\Rightarrow\sqrt{\Delta}=\sqrt{784}=28\)

Vì \(\Delta>0\) nên pt có 2 nghiệm phân biệt

\(\Rightarrow m_1=\dfrac{-32+28}{2\cdot4}=-\dfrac{1}{2}\) (m TM \(m\ge\dfrac{-7}{8}\))

\(m_2=\dfrac{-32-28}{2\cdot4}=-\dfrac{15}{2}\) (m KTM \(m\ge\dfrac{-7}{8}\))

Vậy \(m=-\dfrac{1}{2}\) để 2 nghiệm \(x_1^2;x^2_2\) thỏa mãn \(x_1^2+x^2_2=\dfrac{15}{2}\)

Vậy ...........................