b/ \(A=\left(x_1^2+x_2^2\right)^2-2x_1^2x_2^2+2x_1x_2\)
\(=\left[\left(x_1+x_2\right)^2-2x_1x_2\right]^2-2x_1^2x_2^2+2x_1x_2\)
\(=\left[\left(2m+2\right)^2-2\left(m+1\right)\right]^2-2\left(m+1\right)^2+2\left(m+1\right)\)
\(=\left[4\left(m+1\right)^2-2\left(m+1\right)\right]^2-2\left(m+1\right)^2+2\left(m+1\right)\)
\(=\left[\left(m+1\right)\left(4m+4-2\right)\right]^2-2\left(m+1\right)^2+2\left(m+1\right)\)
\(=\left(m+1\right)^2\left(4m+2\right)^2-2\left(m+1\right)^2+2\left(m+1\right)\)
Đặt m+1=a
\(\Rightarrow a^2\left(4a-2\right)^2-2a^2+2a=A\)
Thui đến đây tắc lun=.=! Bạn tự giải nốt nha:))
\(\Delta'=\left(m+1\right)^2-m-1=m^2+2m+1-m-1=m^2+m\)
Để phương trình có 2 no:
\(\Leftrightarrow\left[{}\begin{matrix}m\ge0\\m\le-1\end{matrix}\right.\)
Theo Vi-ét có:
\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=m+1\end{matrix}\right.\)
Có \(\sqrt{x_1}+\sqrt{x_2}=2\)
\(\Leftrightarrow x_1+x_2+2\sqrt{x_1x_2}=4\)
\(\Leftrightarrow2\left(m+1\right)+2\sqrt{m+1}=4\)
Đặt \(\sqrt{m+1}=a\ge0\Rightarrow m+1=a^2\)
\(\Rightarrow2a^2+2a-4=0\)
\(\Leftrightarrow a=1\)
Thay trở lại có \(\sqrt{m+1}=1\Rightarrow m=0\) (thoả mãn)
