优优班--学霸训练营 > 知识点挑题
全部资源
          排序:
          最新 浏览

          50条信息

            • 1.

              先化简再求值\(\dfrac{2a+2}{a-1}\div \left( a+1 \right)+\dfrac{{{a}^{2}}-1}{{{a}^{2}}-2a+1}\)其中\(a=\sqrt{3}+1\)

            • 2. 计算:
              \((1)\)\(\sqrt{4}-\left| -3 \right|+{{\left( \sqrt{2}-1 \right)}^{0}}\)                 
              \((2)\)\(\dfrac{\sqrt{48}-6}{\sqrt{6}}-6\sqrt{\dfrac{3}{2}}\)
            • 3.

              计算:\(|\sqrt{3}-3|+\sqrt{\dfrac{9}{4}}-\sqrt[3]{27}\) .

            • 4.

              \((1)\)已知\(x= \dfrac{ \sqrt{5}-1}{2}\),\(y= \dfrac{ \sqrt{5}+1}{2}\),求\( \dfrac{y}{x}+ \dfrac{x}{y}\)的值;



               \((2)\)已知\(x\),\(y\)是实数,且\(y < \sqrt{x-2}+ \sqrt{2-x}+ \dfrac{1}{4}\),化简:\( \sqrt{y^{2}-4y+4}-(x-2+ \sqrt{2})^{2}\).

            • 5.

              计算或解方程

              \((1)\)用代入法解方程组:\(\begin{cases}y=3x-2\;\;\;\;① \\ 6x-3y=5\;\;\;②\end{cases} \)         \((2)\)用加减法解方程组:\(\begin{cases} \dfrac{x}{2}+ \dfrac{y}{3}=2 \\ 2x+3y=28\end{cases} \)                    

              \((3)\sqrt[3]{-27}+\sqrt{{{(-3)}^{2}}}-\sqrt[3]{-1}\)    \((4)|-1-\sqrt{2} |-|\sqrt{2}- \sqrt{3} |+|\sqrt{3}- \sqrt{2} |\)

            • 6.

              计算:

              \((1)\sqrt{32}{+}5\sqrt{\dfrac{3}{25}}+4\sqrt{\dfrac{1}{2}}-\sqrt{12}\)           

              \((2){{\left( -\dfrac{a}{b} \right)}^{3}}\div \left( -a{{b}^{3}} \right)\times {{\left( -\dfrac{{{b}^{3}}}{a} \right)}^{2}}\)

            • 7.
              已知\(a{+}b{=-}6{,}{ab}{=}8\),试求\(\sqrt{\dfrac{b}{a}}{+}\sqrt{\dfrac{a}{b}}\)的值
            • 8.

              计算下列各题:

                  \((1)|1-\sqrt{3}|+\sqrt[3]{-27}-2\sqrt{3}+\sqrt{1\dfrac{7}{9}+1}\);

                  \((2)\left| \pi -\sqrt[3]{\dfrac{343}{8}} \right|+\sqrt{{{(3-\pi )}^{2}}}+{{(-2)}^{2}}-\sqrt{\dfrac{1}{4}+2}\) 

            • 9.

              先化简,再求值:\( \dfrac{2a}{{a}^{2}-4}÷\left( \dfrac{{a}^{2}}{a-2}-a\right) \),其中\(a= \sqrt{3}-2 \).

            • 10.

              \((1)\)计算:\(\sqrt{48}\div \sqrt{3}-\sqrt{\dfrac{1}{2}}\times \sqrt{12}\div \sqrt{24}\) 


              \((2)\)若\(a=1+\sqrt{2},b=1-\sqrt{2}\),求\(\sqrt{{{a}^{2}}+{{b}^{2}}-3ab}\)的值 .

              \((3)\)已知\(a+b=-6\)\(ab=8\),求\(\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{a}{b}}\)的值。

            0/40

            进入组卷