题目计算:(1)∣−3∣−(15)−1+4×32−27(1)|-\sqrt{3}|-(\frac{1}{5})^{-1}+4×\frac{\sqrt{3}}{2}-\sqrt{27}(1)∣−3∣−(51)−1+4×23−27；(2)18÷2+(3+2)×(3−2)(2)\sqrt{18}÷\sqrt{2}+(\sqrt{3}+2)×(\sqrt{3}-2)(2)18÷2+(3+2)×(3−2)；(3){2x+3y=13x−2y=−4(3)\left\{\begin{array}{l}2x+3y=13\\ x-2y=-4\end{array}\right.(3){2x+3y=13x−2y=−4；(4){3(x−1)−4(y+1)=−1x2+y3=−2(4)\left\{\begin{array}{l}3(x-1)-4(y+1)=-1\\ \frac{x}{2}+\frac{y}{3}=-2\end{array}\right.(4){3(x−1)−4(y+1)=−12x+3y=−2.知识点：二次根式的性质与化简章节：数与式 / 二次根式
