Ptrình :  \(x^2-7x+10=0\)

Ta có : \(\Delta=\left(-7\right)^2-4.1.10=9>0\)

=> Phương trình có 2 nghiệm phân biệt \(x1\) và \(x2\)

\(x1=\dfrac{-\left(-7\right)+\sqrt{\Delta}}{2.1}=\dfrac{7+\sqrt{9}}{2}=5\)

\(x2=\dfrac{-\left(-7\right)-\sqrt{\Delta}}{2.1}=\dfrac{7-\sqrt{9}}{2}=2\)

Vậy :

A = \(x_1^2+x_2^2+3x_1x_2=5^2+2^2+3.5.2=59\)  

B = .................

.... (có x1 và x2 rồi thik thay vào lak tính đc, cái này bn tự tính nha)

Để phương trình có 2 nghiệm 

\(\Delta'\ge0\Rightarrow\left(-1\right)^2-1.3m\ge0\Leftrightarrow1-3m\ge0\Leftrightarrow1\ge3m\Leftrightarrow\dfrac{1}{3}\ge m\)

Theo hệ thức Vi-et ta có:

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-\left(-2\right)}{1}=2\\x_1x_2=\dfrac{3m}{1}=3m\end{matrix}\right.\)

Ta có: 

\(x_1^2x_2^2=x_1+x_2+7\\ \Leftrightarrow x_1x_2.x_1x_2=x_1+x_2+7\\ \Rightarrow3m.3m=2+7\\ \Leftrightarrow9m^2-9=0\\ \Leftrightarrow\left[{}\begin{matrix}m=-1\left(tm\right)\\m=1\left(ktm\right)\end{matrix}\right.\)

Vậy m = -1

Thai

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

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

1: \(\text{Δ}=\left(2m-2\right)^2-4\left(m-3\right)\)

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

\(=4m^2-12m+16\)

\(=\left(2m-3\right)^2+7>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

2: Theo vi-et, ta được:

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

Ta có: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=x_1x_2\)

\(\Leftrightarrow x_1^2+x_2^2=\left(m-3\right)^2\)

\(\Leftrightarrow\left(2m-2\right)^2-2\left(m-3\right)-\left(m-3\right)^2=0\)

\(\Leftrightarrow4m^2-16m+4-2m+6-m^2+6m-9=0\)

\(\Leftrightarrow3m^2-12m+1=0\)

\(\text{Δ}=\left(-12\right)^2-4\cdot3\cdot1=144-12=132>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{12-2\sqrt{33}}{6}=\dfrac{6-\sqrt{33}}{3}\\x_2=\dfrac{6+\sqrt{33}}{3}\end{matrix}\right.\)

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

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

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 \(m+2=\dfrac{5}{2}\Rightarrow m=\dfrac{1}{2}\)

Cái này phân tích đề ra là lm được bạn nhé

\(\Delta'=m^2-\left(m^2+2m-6\right)=-2m+6\)

a.

Pt có nghiệm khi \(-2m+6\ge0\Rightarrow m\le3\)

b.

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m^2+2m-6\end{matrix}\right.\)

c.

\(x_1x_2=3x_1+3x_2-1\)

\(\Leftrightarrow x_1x_2=3\left(x_1+x_2\right)-1\)

\(\Leftrightarrow m^2+2m-6=3.2m-1\)

\(\Leftrightarrow m^2-4m-5=0\Rightarrow\left[{}\begin{matrix}m=-1\\m=5>3\left(loại\right)\end{matrix}\right.\)

Ta có: \(\Delta=\left(-10\right)^2-4.3.2=100-24=76>0\)

Suy ra pt luôn có 2 nghiệm phân biệt

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{10}{3}\\x_1x_2=\dfrac{2}{3}\end{matrix}\right.\)

\(A=\dfrac{x_1-1}{x_2}+\dfrac{x_2-1}{x_1}-x_1^2x_2^2\\ =\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{x_1x_2}-\left(x_1x_2\right)^2\\ =\dfrac{x_1^2-x_1+x_2^2-x_2}{\dfrac{2}{3}}-\left(\dfrac{2}{3}\right)^2\\ =\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)}{\dfrac{2}{3}}-\dfrac{4}{9}\)

\(=\dfrac{\left(\dfrac{10}{3}\right)^2-2.\dfrac{2}{3}-\dfrac{10}{3}}{\dfrac{2}{3}}-\dfrac{4}{9}\\ =\dfrac{83}{9}\)