Ta có: \(\text{Δ}=\left[-2\left(m+1\right)\right]^2-4\cdot1\cdot\left(m^2+1\right)\)

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

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

=8m

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

hay m>0

Áp dụng hệ thức Vi-et, ta được:

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

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

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

Ta có: \(x_1\cdot x_2=m^2+1\)

\(\Leftrightarrow\dfrac{\left(2m+3\right)\left(2m+1\right)}{4}=m^2+1\)

\(\Leftrightarrow4m^2+2m+6m+3=4m^2+4\)

\(\Leftrightarrow8m=1\)

hay \(m=\dfrac{1}{8}\left(nhận\right)\)

Ta có: \(\Delta\) = m² - 4(m - 1) = m² - 4m + 4 = (m - 2)² \(\ge\) 0

\(\Rightarrow\) x₁ = \(\dfrac{m-\left(m-2\right)}{2}=1\); x₂ = \(\dfrac{m+m-2}{2}=m-1\)

Ta có: |x₁| + |x₂| = 4

\(\Leftrightarrow\) 1 + |m - 1| = 4

\(\Leftrightarrow\) |m - 1| = 3

\(\Leftrightarrow\) \(\left[{}\begin{matrix}m-1=3\\m-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=4\\m=-2\end{matrix}\right.\)

Vậy ...

Chúc bn học tốt!

plz god help me ;-;

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

\(\text{∆}=4\left(m+1\right)^2-16m=4\left(m-1\right)^2\)

để phương trình có 2 nghiệm phân biệt:

\(\Leftrightarrow\left(m-1\right)^2>0\Leftrightarrow m\ne1\)

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

Ta có:

 \(x_1=-3x_2\)

\(\Rightarrow2m=-6\Rightarrow m=-3\left(TM\right)\)

Vậy ...

\(\Delta'=m^2>0\Rightarrow m\ne0\)

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

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

\(\Leftrightarrow2x_1-\left(-m^2+1\right)+x_2=2\)

\(\Leftrightarrow2x_1+x_2=-m^2+3\)

Kết hợp \(x_1+x_2=2\Rightarrow\left\{{}\begin{matrix}2x_1+x_2=-m^2+3\\x_1+x_2=2\end{matrix}\right.\)

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

Thế vào \(x_1x_2=-m^2+1\)

\(\Rightarrow\left(-m^2+1\right)\left(m^2+1\right)=-m^2+1\)

\(\Rightarrow\left[{}\begin{matrix}-m^2+1=0\\m^2+1=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=\pm1\\m=0\left(loại\right)\end{matrix}\right.\)

\(\Delta=\left(-m\right)^2-2.1.\left(m-1\right)\\ =m^2-2m+1\\ =\left(m-1\right)^2\)

Phương trình có hai nghiệm phân biệt :

\(\Leftrightarrow\Delta>0\\ \Rightarrow\left(m-1\right)^2>0\\ \Rightarrow m\ne1\)

Theo vi ét : 

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

\(x^2_1+x^2_2=x_1+x_2\\ \Leftrightarrow x^2_1+x^2_2=m\\ \Leftrightarrow\left(x^2_1+2x_1x_2+x_2^2\right)-2x_1x_2=m\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2-m=0\\ \Leftrightarrow m^2-2\left(m-1\right)-m=0\\ \Leftrightarrow m^2-2m+2-m=0\\ \Leftrightarrow m^2-3m+2=0\\ \Leftrightarrow\left[{}\begin{matrix}m=1\left(loại\right)\\m=2\left(t/m\right)\end{matrix}\right.\)

Vậy \(m=2\)

Phương trình đã cho có hai nghiệm phân biệt khi

\(\Delta'=\left(m+1\right)^2-\left(m^2+2\right)=2m-1>0\Leftrightarrow m>\dfrac{1}{2}\)

Theo định lí Viet: \(x_1+x_2=2m+2;x_1x_2=m^2+2\)

Khi đó \(x_1^3+x_2^3=2x_1x_2\left(x_1+x_2\right)\)

\(\Leftrightarrow\left(x_1+x_2\right)^3-5x_1x_2\left(x_1+x_2\right)=0\)

\(\Leftrightarrow\left(2m+2\right)^3-5\left(m^2+2\right)\left(2m+2\right)=0\)

\(\Leftrightarrow m^3-7m^2-2m+6=0\)

\(\Leftrightarrow\left(m+1\right)\left(m^2-8m+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-1\left(l\right)\\m=4\pm\sqrt{10}\left(tm\right)\end{matrix}\right.\)