Vì phương trình \(\left(x-2a+b-1\right)\left(x+a-2b+1\right)=0\) có hai nghiệm là: \(x=2a-b+1;x=-a+2b-1\).
Ta xét hai trường hợp:
TH1: \(\left\{{}\begin{matrix}2a-b+1=0\\-a+2b-1=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{5}{3}\end{matrix}\right.\).
TH2: \(\left\{{}\begin{matrix}2a-b+1=2\\-a+2b-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
Vậy \(\left(a,b\right)=\left(\dfrac{1}{3};\dfrac{5}{3}\right)\) hoặc \(\left(a,b\right)=\left(1;1\right)\) thì BPT có tập nghiệm là đoạn [0;2].

Các bạn giúp mk với ạ

Nooooooooooo giúp

\(\sqrt{2x-1}< 8-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\8-x\ge0\\2x-1< \left(8-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le8\\x^2-18x+65>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le8\\\left[{}\begin{matrix}x>13\\x< 5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{2}\le x< 5\)

Với $m=-1$ thì PT $f\left(x\right)=0$ có nghiệm $x=1$ (chọn)

Với $m\ne -1$ thì $f\left(x\right)$ là đa thức bậc 2 ẩn $x$

$f\left(x\right)=0$ có nghiệm khi mà ${}^{}={}^{}-2m(m+1)\ge 0$

$\iff -{}^{}-2m\ge 0\iff m(m+2)\le 0$

$\iff -2\le m\le 0$

Tóm lại để $f\left(x\right)=0$ có nghiệm thì $m\in [-2;0]$

Chọn B

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Mình không biết sin lỗi vạn

\(\Leftrightarrow x^2-\left(2m-1\right)x+2m-2=0\) có 2 nghiệm pb \(x_1;x_2\) thỏa mãn \(\left|x_1-x_2\right|=5\)

\(\Delta=\left(2m-1\right)^2-4\left(2m-2\right)=4m^2-12m+9=\left(2m-3\right)^2\)

Pt có 2 nghiệm pb khi \(\left(2m-3\right)^2>0\Rightarrow m\ne\dfrac{3}{2}\)

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

\(\left|x_1-x_2\right|=5\Leftrightarrow\left(x_1-x_2\right)^2=25\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=25\)

\(\Leftrightarrow\left(2m-1\right)^2-4\left(2m-2\right)=25\)

\(\Leftrightarrow\left(2m-3\right)^2=25\)

\(\Rightarrow\left[{}\begin{matrix}2m-3=5\\2m-3=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m=4\\m=-1\end{matrix}\right.\)