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

\(=4m^2+12m+9-16m-8\)

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

Do đó: Phương trình luôn có hai nghiệm

Theo đề, ta có:

\(\left\{{}\begin{matrix}2x_1-5x_2=6\\x_1+x_2=2m+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x_1-5x_2=6\\2x_1+2x_2=4m+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-7x_2=-4m\\2x_1=5x_2+6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{4}{7}m\\2x_1=\dfrac{20}{7}m+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{4}{7}m\\x_1=\dfrac{10}{7}m+3\end{matrix}\right.\)

Theo đề, ta có: \(x_1x_2=4m+2\)

\(\Rightarrow4m+2=\dfrac{40}{49}m^2+\dfrac{12}{7}m\)

\(\Leftrightarrow m^2\cdot\dfrac{40}{49}-\dfrac{16}{7}m-2=0\)

\(\Leftrightarrow40m^2-112m-98=0\)

\(\Leftrightarrow40m^2-140m+28m-98=0\)

=>\(20m\left(2m-7\right)+14\left(2m-7\right)=0\)

=>(2m-7)(20m+14)=0

=>m=7/2 hoặc m=-7/10

Ta có: \(\Delta=\left[-\left(m+3\right)\right]^2-4\left(4m-4\right)=m^2+6m+9-16m+16=\left(m-5\right)^2\ge0\)

=> pt luôn có 2 nghiệm x1, x2

=> \(x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{m+3-m+5}{2}=4\)

  \(x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{m+3+m-5}{2}=m-1\)

Theo bài ra, ta có: \(\sqrt{x_1}+\sqrt{x_2}+x_1x_2=20\)

ĐK: \(x_1\ge0\); \(x_2\ge0\) <=> 4  \(\ge\) 0 và m - 1 \(\ge\)0 <=> m \(\ge\)1

<=> \(\sqrt{4}+\sqrt{m-1}+4\left(m-1\right)=20\)

<=> \(\sqrt{m-1}=22-4m\left(m\le\frac{11}{2}\right)\)

<=> \(m-1=16m^2-176m+484\)

<=> \(16m^2-177m+485=0\)

<=> \(16m^2-80m-97m+485=0\)

<=> \(\left(m-5\right)\left(16m-97\right)=0\)

<=> \(\orbr{\begin{cases}m=5\left(tm\right)\\m=\frac{97}{16}\left(ktm\right)\end{cases}}\)

Vậy ...

\(mx-x-m+2=0\)

\(x\left(m-1\right)=m-2\)

Nếu m=1 ⇒ \(0x=-1\) (vô nghiệm)

Nếu m≠1 ⇒ \(x=\dfrac{m-2}{m-1}\)

Vậy ...