\(1.x^2+2=3x\)
có a+b+c=0(tính Δ cũng cho điểm như vậy)
\(\Rightarrow\left[{}\begin{matrix}x_1=1\\x_2=2\end{matrix}\right.\)
vậy phương trình có nghiệm x=1;x=2
2.\(\left\{{}\begin{matrix}3x=8-2y\\-3y=5-4x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=8\\4x-3y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x+8y=32\\12x-9y=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
vậy hệ phương trình có 1 nghiệm duy nhất (x;y)=(2;1)
\(1,x^2+2=3x\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Delta=b^2-4ac=\left(-3\right)^2-4.2=1>0\)
\(\Rightarrow\) Pt có 2 nghiệm pb \(x_1,x_2\)
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+1}{2}=2\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-1}{2}=1\end{matrix}\right.\)
Vậy \(S=\left\{1;2\right\}\)
\(2,\)\(\left\{{}\begin{matrix}3x=8-2y\\-3y=5-4x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=8\\4x-3y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x+8y=32\\12x-9y=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=8\\17y=17\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2.1=8\\y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy hệ pt có 1 nghiệm duy nhất \(\left(x;y\right)=\left(2;1\right)\)
