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Để pt có 2 nghiệm \(x_1,x_2\) thì \(\Delta'=4\left(m-1\right)^2-3\left(m^2-4m+1\right)=m^2+4m+1\ge0\)
\(\Leftrightarrow\)\(\left(m^2+4m+4\right)-3\ge0\)\(\Leftrightarrow\)\(\left(m+2\right)^2-3\ge0\)
\(\Leftrightarrow\)\(\left(m+2-\sqrt{3}\right)\left(m+2+\sqrt{3}\right)\ge0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}m\ge\sqrt{3}-2\\m\le-\sqrt{3}-2\end{cases}}\)
Ta có : \(\left|x_1-x_2\right|=2\)
\(\Leftrightarrow\)\(\left(x_1-x_2\right)^2=4\)
\(\Leftrightarrow\)\(x_1^2+x_2^2-2x_1x_2=4\)
\(\Leftrightarrow\)\(\left(x_1+x_2\right)^2-4x_1x_2=4\) \(\left(1\right)\)
Theo định lý Vi-et ta có \(\hept{\begin{cases}x_1+x_2=\frac{4\left(1-m\right)}{3}\\x_1x_2=\frac{m^2-4m+1}{3}\end{cases}}\)
\(\left(1\right)\)\(\Leftrightarrow\)\(\left(\frac{4-4m}{3}\right)^2-4\left(\frac{m^2-4m+1}{3}\right)=4\)
\(\Leftrightarrow\)\(\frac{16-32m+16m^2}{9}-\frac{4m^2-16m+4}{3}-4=0\)
\(\Leftrightarrow\)\(\frac{16m^2-32m+16-12m^2+48m-12-36}{9}=0\)
\(\Leftrightarrow\)\(4m^2+16m-32=0\)
\(\Leftrightarrow\)\(\left(m^2+4m+4\right)-12=0\)
\(\Leftrightarrow\)\(\left(m+2\right)^2=12\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}m=2\sqrt{3}-2\left(tm\right)\\m=-2\sqrt{3}-2\left(tm\right)\end{cases}}\)
Vậy để pt có hai nghiệm \(x_1,x_2\) thoả mãn \(\left|x_1-x_2\right|=2\) thì \(\orbr{\begin{cases}m=2\sqrt{3}-2\\m=-2\sqrt{3}-2\end{cases}}\)
chả biết đúng ko nhưng xem thử nha -_-
\(x^2-\left(m+1\right)x+m+4=0\left(1\right)\)
\(\Rightarrow\Delta>0\Leftrightarrow\left(m+1\right)^2-4\left(m+4\right)>0\Leftrightarrow\left[{}\begin{matrix}m< -3\\m>5\end{matrix}\right.\)\(\left(2\right)\)
\(ddkt-thỏa:\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)
\(x1=0\Rightarrow\left(1\right)\Leftrightarrow m=-4\Rightarrow\left(1\right)\Leftrightarrow x^2+3x=0\Leftrightarrow\left[{}\begin{matrix}x1=0\\x2=-3< 0\left(loại\right)\end{matrix}\right.\)
\(x1\ne0\) \(\Rightarrow0< x1< x2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+1>0\\m+4>0\end{matrix}\right.\)\(\Rightarrow m>-1\)\(\left(3\right)\)
\(\left(2\right)\left(3\right)\Rightarrow m>5\)
\(\Rightarrow\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)
\(\Leftrightarrow x1+x2+2\sqrt{x1x2}=12\Leftrightarrow m+1+2\sqrt{m+4}=12\)
\(\Leftrightarrow m+4+2\sqrt{m+4}-15=0\)
\(đặt:\sqrt{m+4}=t>5\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=-5\left(ktm\right)\\t=3\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow m\in\phi\)
Để pt có 2 nghiệm pb
\(\left(m+1\right)^2-4\left(m+4\right)=m^2+2m+1-4m-16\)
\(=m^2-2m-15>0\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1x_2=m+4\end{matrix}\right.\)
Ta có : \(\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=12\Leftrightarrow x_1+2\sqrt{x_1x_2}+x_2=12\)
Thay vào ta được \(m+1+2\sqrt{m+4}=12\Leftrightarrow2\sqrt{m+4}=11-m\)đk : m >= -4
\(\Leftrightarrow4\left(m+4\right)=121-22m+m^2\Leftrightarrow m^2-26m+105=0\)
\(\Leftrightarrow m=21\left(ktm\right);m=5\left(ktm\right)\)
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 ...