a) \(\Delta\)=(m-3)²-4.1.(2m-11)=m²-14m+53=(m-7)²+4\(\ge\)4.

\(\Rightarrow\) Phương trình đã cho luôn có hai nghiệm phân biệt với mọi m.

b) Từ ycđb, ta có: x₁²+x₂²=4² \(\Leftrightarrow\) (x₁+x₂)²-2x₁x₂=16 \(\Leftrightarrow\) (m-3)²-2(2m-11)=16 \(\Leftrightarrow\) m²-10m+15=0 \(\Leftrightarrow\) \(m=5\pm\sqrt{10}\).

\(a+b+c=1-\left(2m+1\right)+2m=0\)

\(\Rightarrow\) Phương trình có 2 nghiệm \(x=1\) ; \(x=2m\)

Để pt có 2 nghiệm pb \(\Rightarrow2m\ne1\Rightarrow m\ne\dfrac{1}{2}\)

\(\left|x_1^2-x_2^2\right|=35\)

\(\Leftrightarrow\left|4m^2-1\right|=35\)

\(\Leftrightarrow\left[{}\begin{matrix}4m^2-1=35\\4m^2-1=-35\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m^2=9\\m^2=-\dfrac{17}{2}\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m=3\\m=-3< 0\left(loại\right)\end{matrix}\right.\)

Ta có \(\Delta=\left(m+2\right)^2-4\left(m+8\right)>0\)

<=> \(m^2-28>0\)

<=> \(\orbr{\begin{cases}m>\sqrt{28}\\m< -\sqrt{28}\end{cases}}\)

Áp dụng hệ thức vi-et ta có

\(\hept{\begin{cases}x_1+x_2=m+2\\x_1x_2=m+8\end{cases}}\)

=> \(x_1+x_2-x_1x_2+6=0\)

Mà \(x_1^3=x_2\)

=> \(x_1^3+x_1-x_1^4+6=0\)

<=> \(\)\(x_1=2\)

=> m=8(thỏa mãn ĐK)

Vậy m=8

bạn ơi sao suy ra đc x1=2 lun v

1.

Yêu cầu bài toán thỏa mãn khi:

\(\left\{{}\begin{matrix}\Delta=25-12m>0\\x_1^2+x_2^2< 17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{25}{12}\\\left(x_1+x_2\right)^2-2x_1x_2< 17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{25}{12}\\\left(2m-3\right)^2-2\left(m^2-4\right)< 17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{25}{12}\\2m^2-12m< 0\end{matrix}\right.\)

\(\Leftrightarrow0< m< \dfrac{25}{12}\)

3.

Yêu cầu bài toán thỏa mãn khi:

\(\left\{{}\begin{matrix}\Delta'=11-m>0\\x_1+x_2>0\\x_1x_2>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 11\\6>0\\m-2>0\end{matrix}\right.\)

\(\Leftrightarrow2< m< 11\)

a thay vào mà tính, dễ rồi nên mình ko làm nữa nhé

b, Để phương trình  có 2 nghiệm phân biệt thì delta > 0 

hay \(4m^2-4\left(m-2\right)\left(m-4\right)=4m^2-4\left(m^2-6m+8\right)=6m-8>0\)

\(\Leftrightarrow-8>-6m\Leftrightarrow m>\dfrac{4}{3}\)

c, Theo Vi et ta có : \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{2m}{m-4}\\x_1x_2=\dfrac{c}{a}=\dfrac{m-2}{m-4}\end{matrix}\right.\)

Lại có: \(\left(x_1+x_2\right)^2=\dfrac{4m^2}{\left(m-4\right)^2}\Rightarrow x_1^2+x_2^2=\dfrac{4m^2}{\left(m-4\right)^2}-2x_1x_2\)

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

\(=\dfrac{4m^2-2m^2+12m-16}{\left(m-4\right)^2}=\dfrac{2m^2+12m-16}{\left(m-4\right)^2}\)

\(\text{Δ}=\left(-5\right)^2-4\cdot1\cdot\left(m+2\right)\)

\(=25-4m-8=-4m+17\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>-4m+17>0

=>-4m>-17

=>\(m< \dfrac{17}{4}\)

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-5\right)}{1}=5\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{m+2}{1}=m+2\end{matrix}\right.\)

\(P=x_1^2\cdot x_2+x_1\cdot x_2^2-x_1^2\cdot x_2^2-4\)

\(=x_1x_2\left(x_1+x_2\right)-\left(x_1x_2\right)^2-4\)

\(=5\left(m+2\right)-\left(m+2\right)^2-4\)

\(=5m+10-m^2-4m-4-4\)

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

\(=-\left(m^2-m-2\right)\)

\(=-\left(m^2-m+\dfrac{1}{4}-\dfrac{9}{4}\right)\)

\(=-\left(m-\dfrac{1}{2}\right)^2+\dfrac{9}{4}< =\dfrac{9}{4}\forall m\)

Dấu '=' xảy ra khi \(m=\dfrac{1}{2}\)

\(\Delta=25-4\left(m+2\right)=17-4m>0\Rightarrow m< \dfrac{17}{4}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=m+2\end{matrix}\right.\)

\(P=x_1x_2\left(x_1+x_2\right)-\left(x_1x_2\right)^2-4\)

\(=5\left(m+2\right)-\left(m+2\right)^2-4\)

\(=-\left[\left(m+2\right)-\dfrac{5}{2}\right]^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

\(P_{max}=\dfrac{9}{4}\) khi \(m+2=\dfrac{5}{2}\Rightarrow m=\dfrac{1}{2}\)

`1)`

$a\big)\Delta=7^2-5.4.1=29>0\to$ PT có 2 nghiệm pb

$b\big)$

Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{7}{5}\\x_1x_2=\dfrac{1}{5}\end{matrix}\right.\)

\(A=\left(x_1-\dfrac{7}{5}\right)x_1+\dfrac{1}{25x_2^2}+x_2^2\\ \Rightarrow A=\left(x_1-x_1-x_2\right)x_1+\left(\dfrac{1}{5}\right)^2\cdot\dfrac{1}{x_2^2}+x_2^2\\ \Rightarrow A=-x_1x_2+\left(x_1x_2\right)^2\cdot\dfrac{1}{x_2^2}+x_2^2\)

\(\Rightarrow A=-x_1x_2+x_1^2+x_2^2\\ \Rightarrow A=\left(x_1+x_2\right)^2-3x_1x_2\\ \Rightarrow A=\left(\dfrac{7}{5}\right)^2-3\cdot\dfrac{1}{5}=\dfrac{34}{25}\)

​\(\Delta'=m^2+1\Rightarrow\left\{{}\begin{matrix}x_1=m+1+\sqrt{m^2+1}\\x_2=m+1-\sqrt{m^2+1}\end{matrix}\right.\)

(Do \(m+1-\sqrt{m^2+1}< \sqrt{m^2+1}+1-\sqrt{m^2+1}< 4\) nên nó ko thể là nghiệm \(x_1\))

Từ điều kiện \(x_1\ge4\Rightarrow m+1+\sqrt{m^2+1}\ge4\Rightarrow\sqrt{m^2+1}\ge3-m\)

\(\Rightarrow\left[{}\begin{matrix}m\ge3\\\left\{{}\begin{matrix}m< 3\\m^2+1\ge m^2-6m+9\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m\ge\dfrac{4}{3}\)

​​Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m\end{matrix}\right.\)

\(x_1^2=9x_2+10\Leftrightarrow x_1\left(x_1+x_2\right)-x_1x_2=9x_2+10\)

\(\Leftrightarrow2\left(m+1\right)x_1-2m=9x_2+10\)

\(\Leftrightarrow2\left(m+1\right)x_1-2m=9\left(2\left(m+1\right)-x_1\right)+10\)

\(\Leftrightarrow\left(2m+11\right)x_1=20m+28\Rightarrow x_1=\dfrac{20m+28}{2m+11}\) 

\(\Rightarrow x_2=2\left(m+1\right)-x_1=\dfrac{4m^2+6m-6}{2m+11}\)

Thế vào \(x_1x_2=2m\)

\(\Rightarrow\left(\dfrac{20m+28}{2m+11}\right)\left(\dfrac{4m^2+6m-6}{2m+11}\right)=2m\)

\(\Leftrightarrow\left(3m-4\right)\left(12m^2+40m+21\right)=0\)

\(\Leftrightarrow m=\dfrac{4}{3}\) (do \(12m^2+40m+21>0;\forall m\ge\dfrac{4}{3}\))

Lời giải:

Để PT có 2 nghiệm phân biệt $x_1,x_2$ thì:

\(\Delta'=(m+2)^2-(m^2+m+3)>0\)

\(\Leftrightarrow 3m+1>0\Leftrightarrow m> \frac{-1}{3}\)

Áp dụng định lý Vi-et: \(\left\{\begin{matrix} x_1+x_2=2(m+2)\\ x_1x_2=m^2+m+3\end{matrix}\right.\)

\(x_1x_2=m^2+m+3=(m+\frac{1}{2})^2+\frac{11}{4}\neq 0, \forall m>\frac{-1}{3}\) nên $x_1,x_2\neq 0$ với mọi \(m> \frac{-1}{3}\).

Khi đó:

\(\frac{x_1}{x_2}+\frac{x_2}{x_1}=1\)

\(\Leftrightarrow \frac{x_1^2+x_2^2}{x_1x_2}=4\)

\(\Leftrightarrow \frac{(x_1+x_2)^2-2x_1x_2}{x_1x_2}=4\)

\(\Leftrightarrow \frac{(x_1+x_2)^2}{x_1x_2}=6\Rightarrow (x_1+x_2)^2=6x_1x_2\)

\(\Leftrightarrow 4(m+2)^2=6(m^2+m+3)\)

\(\Leftrightarrow 2m^2-10m+2=0\)

\(\Leftrightarrow m=\frac{5\pm \sqrt{21}}{2}\) (thỏa mãn)