Đáp án:
$1)\\ a) x=1\\ b) x=\pm 2\\ 2)\\ a)\left\{\begin{array}{l} x=1\\ y=2\end{array} \right.\\ b)m=\dfrac{-1 \pm \sqrt{19}}{2}$
Giải thích các bước giải:
$1)\\ a)5(x+1)=3x+7\\ \Leftrightarrow 5x+5=3x+7\\ \Leftrightarrow 2x=2\\ \Leftrightarrow x=1\\ b)x^4-x^2-12=0\\ \Leftrightarrow x^4-4x^2+3x^2-12=0\\ \Leftrightarrow x^2(x^2-4)+3(x^2-4)=0\\ \Leftrightarrow (x^2+4)(x^2-4)=0\\ \Leftrightarrow (x^2+4)(x-2)(x+2)=0\\ \Leftrightarrow x=\pm 2\\ 2)\\ \left\{\begin{array}{l} 3x-y=2m-1\\ x+2y=3m+2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 6x-2y=4m-2\\ x+2y=3m+2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 6x-2y=4m-2\\ x+2y+6x-2y=3m+2+4m-2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 3x-y=2m-1\\ 7x=7m \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=m\\ y=m+1\end{array} \right.(*)\\ a)m=1, (*) \Leftrightarrow \left\{\begin{array}{l} x=1\\ y=2\end{array} \right.\\ b)x^2+y^2=10\\ \Leftrightarrow m^2+(m+1)^2=10\\ \Leftrightarrow m^2+m^2+2m+1=10\\ \Leftrightarrow 2m^2+2m-9=0\\ \Leftrightarrow m=\dfrac{-1 \pm \sqrt{19}}{2}$
