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            • 1.

              计算

              \((1){{(2\sqrt{5}-\sqrt{2})}^{0}}+|2-\sqrt{5}|+{{(-1)}^{2018}}-\dfrac{1}{3}\times \sqrt{45}\).

              \((2){{(\pi -3)}^{0}}+|\sqrt{2}-2|+\sqrt{10}\div \sqrt{5}-{{1}^{-2}}\).

              \((3)\dfrac{2}{3}\sqrt{9x}+6\sqrt{\dfrac{x}{4}}-2x\sqrt{\dfrac{1}{x}}(x > 0)\).

            • 2.

              \((1)\)计算:\(|2−\tan ⁡{60}^{∘}|−(π−3.14{)}^{0}+(− \dfrac{1}{2}{)}^{−2}+ \dfrac{1}{2} \sqrt{12} \).

              \((2)\)解方程:\((x{-}1)(x +3){=}12\).
            • 3.

              计算:

               \((1)\sqrt{81} +\sqrt[3]{-27} +\sqrt{{\left(- \dfrac{2}{3}\right)}^{2}} \)    

              \((2)3\sqrt{2} -|\sqrt{3} -\sqrt{2} |.\)

            • 4. 已知\(|x-2|+1-2y+y^{2}=0\),求\(x^{2}+(2xy-3y^{2})- 2(x^{2}+xy-2y^{2})\)的值.
            • 5.

              计算:\({|-}5{|+}(\pi{-}3{.}1)^{0}{-}(\dfrac{1}{2})^{-1}{+}\sqrt{4}\);

            • 6.

              解答题

              \((1)\)解方程:\(9(x-2)^{2}=4(x+1)^{2}\)

              \((2)|1-\sqrt{3}|-2\cos 30{}^\circ +{{(-\dfrac{1}{2})}^{0}}\times {{(-1)}^{2013}}\)

            • 7.

              计算:\(\sqrt{18}+(π-3)^{0}-(-\sqrt{5})^{-2}+|2\sqrt{2}-3|\)

            • 8.

              \((1)\)计算:\((– \sqrt[]{3})^{0}–\left| –3 \right|+ (–1)^{2016}+ ( \dfrac{1}{2})^{–1}\)


              \((2)\)解不等式组:\(\begin{cases}x-1 > 2 \\ x+2 < 4x-1\end{cases} \).

            • 9.

              计算:

              \((1) \dfrac{{x}^{2}}{x-y}- \dfrac{{y}^{2}}{x-y} \)                    \((2)\) .\( \dfrac{2{b}^{2}}{a+b}-a+b \)


              \((3){\left( \dfrac{y}{6{x}^{2}}\right)}^{2}÷{\left(- \dfrac{{y}^{2}}{4x}\right)}^{2} \)             \((4) \sqrt{6}-\left(2 \sqrt{ \dfrac{3}{2}}-3 \sqrt{ \dfrac{2}{3}}\right)- \sqrt{24} \)


              \((5)\left(2 \sqrt{5}+3 \sqrt{2}\right)\left(2 \sqrt{5}-3 \sqrt{2}\right) _{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}(6)\left|- \sqrt{2}\right|- \sqrt{8}+{\left(1- \sqrt{3}\right)}^{0}+ \dfrac{1}{ \sqrt{2}+ \sqrt{3}} \)

            • 10.

              已知\( \dfrac{2}{3} (m+4)x^{|m|-3}+6 > 0\)是关于\(x\)的一元一次不等式,则\(m=\)  

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