\(\Delta=m^2-4\left(m-1\right)=\left(m-2\right)^2\ge0;\forall m\) nên pt luôn có 2 nghiệm

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-1\end{matrix}\right.\)

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

\(=\dfrac{2\left(m-1\right)+3}{m^2+2}=\dfrac{2m+1}{m^2+2}=\dfrac{4m+2}{2\left(m^2+2\right)}=\dfrac{m^2+4m+4-\left(m^2+2\right)}{2\left(m^2+2\right)}\)

\(=\dfrac{\left(m+2\right)^2}{2\left(m^2+2\right)}-\dfrac{1}{2}\ge-\dfrac{1}{2}\)

Vậy \(B_{min}=-\dfrac{1}{2}\)

1.

\(a+b+c=0\) nên pt luôn có 2 nghiệm

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

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

\(A=\dfrac{m^2+2-\left(m^2-2m+1\right)}{m^2+2}=1-\dfrac{\left(m-1\right)^2}{m^2+2}\le1\)

Dấu "=" xảy ra khi \(m=1\)

2.

\(\Delta=m^2-4\left(m-2\right)=\left(m-2\right)^2+4>0;\forall m\) nên pt luôn có 2 nghiệm pb

Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-2\end{matrix}\right.\)

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

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

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

\(\Rightarrow-m^2=-4\Rightarrow m=\pm2\)

Giả sử ta định m sao cho pt \(x^2-mx+m-1=0\left(1\right)\) luôn có nghiệm.

Theo định lí Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-1\end{matrix}\right.\)

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

\(\Rightarrow C\left(m^2+2\right)=2m+1\Rightarrow Cm^2-2m+\left(2C+1\right)=0\left(2\right)\)

Coi phương trình (2) là phương trình ẩn m tham số C, ta có:

\(\Delta'=1^2-C.\left(2C+1\right)=-2C^2-C+1\)

Để phương trình (2) có nghiệm thì:

\(\Delta'\ge0\Rightarrow-2C^2-C+1\ge0\)

\(\Leftrightarrow\left(2C-1\right)\left(C+1\right)\le0\)

\(\Leftrightarrow-1\le C\le\dfrac{1}{2}\)

Vậy \(MinC=-1;MaxC=\dfrac{1}{2}\)

\(\Delta'=m^2-2\left(m^2-2\right)=4-m^2\ge0\Rightarrow-2\le m\le2\)

Khi đó ta có \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=\frac{m^2-2}{2}\end{matrix}\right.\)

\(A=\frac{2x_1x_2+3}{x_1^2+x_2^2+2x_1x_2+2}=\frac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=\frac{m^2+1}{m^2+2}=1-\frac{1}{m^2+2}\)

Do \(0\le m^2\le4\Rightarrow\frac{1}{6}\le\frac{1}{m^2+2}\le\frac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}A_{min}=1-\frac{1}{2}=\frac{1}{2}\Rightarrow m=0\\A_{max}=1-\frac{1}{6}=\frac{5}{6}\Rightarrow m=\pm2\end{matrix}\right.\)

Bạn vui lòng đối chiếu đề bạn đang có giúp mình ở hai chữ "x" mình in đậm nhé! Mình sẽ hỗ trợ nhanh nhất có thể!

Đề: Cho phương trình: xx²−(m−3)x−m=3 (1).

a) Chứng minh phương trình có hai nghiệm phân biệt.

b) Tìm m đề hai nghiệm x₁, x₂ của phương trình thoả mãn hệ thức: 3x(x₁+x₂)−x₁x₂≥5.

Xin cảm ơn!

b) phương trình có 2 nghiệm  \(\Leftrightarrow\Delta'\ge0\)

\(\Leftrightarrow\left(m-1\right)^2-\left(m-1\right)\left(m+3\right)\ge0\)

\(\Leftrightarrow m^2-2m+1-m^2-3m+m+3\ge0\)

\(\Leftrightarrow-4m+4\ge0\)

\(\Leftrightarrow m\le1\)

Ta có: \(x_1^2+x_1x_2+x_2^2=1\)

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

Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=m+3\end{matrix}\right.\)

\(\Leftrightarrow\left[-2\left(m-1\right)^2\right]-2\left(m+3\right)=1\)

\(\Leftrightarrow4m^2-8m+4-2m-6-1=0\)

\(\Leftrightarrow4m^2-10m-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m_1=\dfrac{5+\sqrt{37}}{4}\left(ktm\right)\\m_2=\dfrac{5-\sqrt{37}}{4}\left(tm\right)\end{matrix}\right.\Rightarrow m=\dfrac{5-\sqrt{37}}{4}\)

Câu c:

xét pt \(x^2-mx+m-1=0\)  \(\left(1\right)\)

xó \(\Delta=\left(-m\right)^2-4\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2>0\forall m\ne2\)

\(\Rightarrow pt\)  (1) có 2 nghiệm phân biệt \(x_1,x_2\forall m\ne2\)

ta có vi -ét \(\hept{\begin{cases}x_1+x_2=m\\x_1.x_2=m-1\end{cases}}\)

theo bài ra \(\left|x_1\right|+\left|x_2\right|=6\)

\(\Leftrightarrow\left(\left|x_1\right|+\left|x_2\right|\right)^2=36\)

\(\Leftrightarrow x_1^2+x_2^2+2\left|x_1.x_2\right|=36\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=36\)

\(\Leftrightarrow m^2-2\left(m-1\right)+2\left|m-1\right|=36\)

nếu \(m-1< 0\Rightarrow m^2-4m-32=0\)  ta tìm được \(m=8\left(loai\right)\); \(m=-4\left(TM\right)\)

nếu \(m-1\ge0\Rightarrow m^2=36\Rightarrow m=6\left(TM\right);m=-6\left(loai\right)\)

vậy \(m=-4;m=6\)  là các giá trị cần tìm 

b) \(P=\frac{2x_1x_2+3}{\left(x_1+x_2\right)^2-2x_1x_2+2x_1x_2+2}\)

\(P=\frac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=\frac{2\left(m-1\right)+3}{m^2+2}\)

\(P=\frac{2m-2+3}{m^2+2}=\frac{2m+1}{m^2+2}\)

vậy \(P=\frac{2m+1}{m^2+2}\)

c) \(P=\frac{2m+1}{m^2+2}=\frac{m^2+2-m^2+2m-1}{m^2+2}=1-\frac{m^2-2m+1}{m^2+2}\)

\(P=1-\frac{\left(m-1\right)^2}{m^2+2}\le1\)

dấu \("="\)  xảy ra \(\Leftrightarrow m-1=0\Leftrightarrow m=1\)

vậy \(MAX\)   \(P=1\Leftrightarrow m=1\)

\(P=\frac{2m+1}{m^2+2}=\frac{4m+2}{2\left(m^2+2\right)}=\frac{m^2+4m+4-m^2-2}{2\left(m^2+2\right)}\)

\(P=\frac{\left(m+2\right)^2-m^2-2}{2\left(m^2+2\right)}=\frac{\left(m+2\right)^2}{2\left(m^2+2\right)}-\frac{m^2+2}{2\left(m^2+2\right)}\)

\(P=\frac{\left(m+2\right)^2}{2\left(m^2+2\right)}-\frac{1}{2}\ge\frac{1}{2}\)

dấu \("="\)  xảy ra \(\Leftrightarrow m+2=0\Leftrightarrow m=-2\)

vậy \(MIN\)   \(P=\frac{1}{2}\Leftrightarrow m=-2\)