1: \(\left\{{}\begin{matrix}3x+y=1\\x-2y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x+2y=2\\x-2y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x+2y+x-2y=2+5=7\\x-2y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x=7\\2y=x-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\2y=1-5=-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
2:
a: Thay m=2 vào phương trình, ta được:
\(x^2-2\left(2+2\right)x+2^2+7=0\)
=>\(x^2-8x+11=0\)
=>\(x^2-8x+16=5\)
=>\(\left(x-4\right)^2=5\)
=>\(x-4=\pm\sqrt{5}\)
=>\(x=4\pm\sqrt{5}\)
b: \(\text{Δ}=\left[-2\left(m+2\right)\right]^2-4\left(m^2+7\right)\)
\(=4\left(m^2+4m+4\right)-4\left(m^2+7\right)\)
\(=4\left(4m-3\right)\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>4m-3>0
=>m>3/4
Theo Vi-et, ta có:
\(x_1+x_2=-\dfrac{b}{a}=2m+4;x_1x_2=\dfrac{c}{a}=m^2+7\)
\(x_1^2+x_2^2=x_1x_2+12\)
=>\(\left(x_1+x_2\right)^2-3x_1x_2=12\)
=>\(\left(2m+4\right)^2-3\cdot\left(m^2+7\right)-12=0\)
=>\(4m^2+16m+16-3m^2-21-12=0\)
=>\(m^2+16m-17=0\)
=>(m+17)(m-1)=0
=>\(\left[{}\begin{matrix}m=-17\left(loại\right)\\m=1\left(nhận\right)\end{matrix}\right.\)
