题目解下列方程:(1)(x+1)2−4=0(1)\left(x+1\right)^{2}-4=0(1)(x+1)2−4=0；(2)y(y−1)=2(y−1)(2)y\left(y-1\right)=2\left(y-1\right)(2)y(y−1)=2(y−1)；(3)2x2−3x−1=0(3)2x^{2}-3x-1=0(3)2x2−3x−1=0；(4)(x+3)(x−1)=−4(4)\left(x+3\right)\left(x-1\right)=-4(4)(x+3)(x−1)=−4.知识点：解二元一次方程组——代入消元法章节：方程与不等式 / 二元一次方程组
