1:
a: \(\left\{{}\begin{matrix}y-2x=5\\2y-x=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2y-4x=10\\2y-x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y-4x-2y+x=10-1\\2y-x=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-3x=9\\2y=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\2y=-3+1=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)
b: \(x^6+7x^3=8\)
=>\(x^6+7x^3-8=0\)
=>\(\left(x^3+8\right)\left(x^3-1\right)=0\)
=>\(\left[{}\begin{matrix}x^3+8=0\\x^3-1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
2:
a: \(\text{Δ}=\left[-\left(2m-3\right)\right]^2-4\cdot1\cdot\left(m^2-3m\right)\)
\(=4m^2-12m+9-4m^2+12m=9>0\)
=>Phương trình luôn có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x=\dfrac{2m-3-\sqrt{9}}{2}=\dfrac{2m-3-3}{2}=m-3\\x=\dfrac{2m-3+3}{2}=m\end{matrix}\right.\)
b: Để phương trình có hai nghiệm trái dấu thì a*c<0
=>\(m^2-3m< 0\)
=>m(m-3)<0
=>0<m<3
