Question

[Kiên Là Tôi]

Cho phương trình: \(x^2+x\left(m+1\right)+m=2\)

1) Chứng minh rằng với mọi m phương trình luôn có 2 nghiệm phân biệt

2) Tìm m sao cho phương trình có 2 nghiệm thỏa mãn: \(\dfrac{2x_1-1}{x_2}+\dfrac{2x_2-1}{x_1}=x_1x_2+\dfrac{55}{x_1x_2}\)

Answer 1

[Kuro Kazuya]

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

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

\(\Delta=b^2-4ac\)

\(\Delta=\left(m-1\right)^2+8\ge8\) \(\forall m\in R\)

\(\Rightarrow\) đpcm

2)

Theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=-m-1\\x_1x_2=m-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x_1+x_2\right)^2=\left(-m-1\right)^2\\2x_1x_2=2m-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+2x_1x_2+x^2_2=m^2+2m+1\\2x_1x_2=2m-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+x^2_2+2m-4=m^2+2m+1\\2x_1x_2=2m-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+x^2_2=\left(m^2+2m+1\right)-\left(2m-4\right)\\2x_2x_1=2m-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+x^2_2=m^2+5\\x_1x_2=m-2\end{matrix}\right.\)

Theo yêu cầu đề bài \(\dfrac{2x_1-1}{x_2}+\dfrac{2x_2-1}{x_1}=x_1x_2+\dfrac{55}{x_1x_2}\)

\(\Leftrightarrow\dfrac{2\left(x^2_1+x^2_2\right)-\left(x_1+x_2\right)}{x_1x_2}=x_1x_2+\dfrac{55}{x_1x_2}\)

\(\Leftrightarrow\dfrac{2\left(m^2+5\right)-\left(-m-1\right)}{m-2}=m-2+\dfrac{55}{m-2}\)

\(\Leftrightarrow\dfrac{2m^2+m+11}{m-2}=\dfrac{\left(m-2\right)^2+55}{m-2}\)

\(\Leftrightarrow2m^2+m+11=\left(m-2\right)^2+55\) ( điều kiện \(m\ne2\) )

\(\Leftrightarrow m^2+5m-48=0\)

\(\Delta=b^2-4ac\)

\(\Rightarrow\Delta=217\)

\(\Rightarrow m_{1,2}=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-5\pm\sqrt{217}}{2}\)

Answer 2

[ngonhuminh]

1)Ta có: \(\Delta_{x,m}=\left(m+1\right)^2-4\left(m-2\right)=\left(m-1\right)^2+8\ge8>0=>dpcm\)

2)Gọi hai nghiệm là a,b cho dẽ viết:

Đặt \(A=\dfrac{2a-1}{b}+\dfrac{2b-1}{a}-ab-\dfrac{55}{ab}\) rút gọn A trước

\(\Leftrightarrow\dfrac{2\left(a^2+b^2\right)-\left(a+b\right)-\left(ab\right)^2-55}{ab}\)

\(A=\dfrac{2\left(a+b\right)^2-\left(a+b\right)-\left[\left(ab\right)^2+4ab+4\right]-51}{ab}\)

\(A=\dfrac{\left(a+b\right)\left[2\left(a+b\right)-1\right]-\left[\left(ab\right)+2\right]^2-51}{ab}\) (1)

Thay Vi_et vào (1) \(\left\{{}\begin{matrix}a+b=-\left(m+1\right)\\ab=m-2\end{matrix}\right.\)

\(A=\dfrac{\left(m+1\right)\left(2m+3\right)-m^2-51}{m-2}=\dfrac{m^2+5m-48}{m-2}\)

\(A=0\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\m^2+5m-48=0\end{matrix}\right.\)

\(\Delta_m=25+4.48=217\) \(\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\\left[{}\begin{matrix}m_2=\dfrac{-5-\sqrt{217}}{2}\ne2\\m_2=\dfrac{-5+\sqrt{217}}{2}\ne2\end{matrix}\right.\end{matrix}\right.\)

Kết luận: \(m_{1,2}=\dfrac{-5\pm\sqrt{217}}{2}\)