题目计算：(1)(-1)2-(2-3)+∣3-2∣(1)\sqrt{(\text{-}1)^{2}}\text{-}(2\text{-}\sqrt{3})\text{+}\left|\sqrt{3}\text{-}2\right|(1)(-1)2-(2-3)+3-2 (2)-22×(12)2+25+-643÷∣-2∣(2)\text{-}2^{2}\text{×}(\dfrac{1}{2})^{2}\text{+}\sqrt{25}\text{+}\sqrt[3]{\text{-}64}\text{÷}\left|\text{-}2\right|(2)-22×(21)2+25+3-64÷∣-2∣(3){3x+4y=165x−6y=33(3)\begin{cases}3x+4y=16\\5x-6y=33\end{cases}(3){3x+4y=165x−6y=33 (4){2(x−1)−3(y+1)=12x2+y3=1(4)\begin{cases}2(x-1)-3(y+1)=12\\\dfrac{x}{2}+\dfrac{y}{3}=1\end{cases}(4){2(x−1)−3(y+1)=122x+3y=1知识点：解二元一次方程组——加减消元法章节：方程与不等式 / 二元一次方程组
