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.
              下列计算正确的是\((\)  \()\)
              A.\( \sqrt {20}=2 \sqrt {10}\)
              B.\( \sqrt {2}⋅ \sqrt {3}= \sqrt {6}\)
              C.\( \sqrt {4}- \sqrt {2}= \sqrt {2}\)
              D.\( \sqrt {(-3)^{2}}=-3\)
            • 2.

              \((1)\)计算:\(\left( 4\sqrt{6}{-}6\sqrt{2} \right)\div 2\sqrt{2}=\)__________.

              \((2)\)如图,在\(□\)\(ABCD\)中,\(AB=4\),\(BC=7\),\(∠ABC\)的平分线\(BE\)交\(AD\)于点\(E\),则\(DE=\)____________\(.\)      

                                    

              \((3)\)已知直线\(y=(2-3m)x\)经过点\(A(x_{1},y_{1})\)、\(B(x_{2},y_{2})\),当\(x_{1} < x_{2}\)时,有\(y_{1} > y_{2}\),则\(m\)的取值范围是__________\(.\)    

              \((4)\)已知一个直角三角形的两边长分别为\(4\)和\(3\),则它的面积为_____________.

              \((5)\)如图,四边形\(OABC\)为矩形,点\(A\),\(C\)分别在\(x\)轴和\(y\)轴上,连接\(AC\),点\(B\)的坐标为\((4,3)\),\(∠CAO\)的平分线与\(y\)轴相交于点\(D\),则点\(D\)的坐标为_________.

                  

              \((6)\)如图,\(□\)\(ABCD\)中,\(AB=2\),\(BC=4\),\(∠B=60^{\circ}\),点\(P\)是四边形上的一个动点,则当\(\triangle PBC\)为直角三角形时,\(BP\)的长为_____________.

            • 3.

              \((1)\)计算:\({{(3-2\sqrt{3})}^{2}}={ }\!\!\_\!\!{ }\!\!\_\!\!{ }\!\!\_\!\!{ }\!\!\_\!\!{ }\!\!\_\!\!{ }\!\!\_\!\!{ }\!\!\_\!\!{ }\!\!\_\!\!{ }\).

              \((2)\)若\(a\)、\(b\)、\(c\)为三角形的三边,且\(a\)、\(b\)满足\(\sqrt{a-10}+{{(b-2)}^{2}}=0\),第三边\(c\)为偶数,则\(c=\)________.

              \((3)\)已知\(x=-1\)是一元二次方程\(x^{2}+ax+b=0\)的一个根,则\(a^{2}-2ab+b^{2}\)的值为________.

              \((4)\)若矩形的对角线长为\(8cm\),两条对角线的一个交角为\(60^{\circ}\),则矩形的面积为________\(cm^{2}\).

              \((5)\)如图,四边形\(ABCD\)是菱形\(.\)对角线\(AC=8cm\),\(DB=6cm\),\(DH⊥AB\)与点\(H.\)则\(DH=\)________\(cm\).

              \((6)\)一元二次方程\((2x+1)^{2}-81=0\)的根是________.

              \((7)\)如图,正方形\(ABCD\)的边长为\(3\),点\(E\)在边\(AB\)上,且\(BE=1.\)若点\(P\)在对角线\(BD\)上移动,则\(PA+PE\)的最小值是________.

              \((8)\)观察下列各式:\(①\sqrt{1+\dfrac{1}{3}}=2\sqrt{\dfrac{1}{3}}\),\(②\sqrt{2+\dfrac{1}{4}}=3\sqrt{\dfrac{1}{4}}\),\(③\sqrt{3+\dfrac{1}{5}}=4\sqrt{\dfrac{1}{5}}\),\(…\),根据你所发现的规律写出第\(n(n\geqslant 1)\)个等式来:________.

            • 4.
              \((1) \sqrt {8}+2 \sqrt {3}-( \sqrt {27}- \sqrt {2})\)
              \((2)(2 \sqrt {5}-5 \sqrt {2})(-2 \sqrt {5}-5 \sqrt {2})-( \sqrt {5}- \sqrt {2})^{2}\).
            • 5.
              观察下列一组式的变形过程,然后回答问题:
              例\(1\):\( \dfrac {1}{ \sqrt {2}+1}= \dfrac { \sqrt {2}-1}{( \sqrt {2}+1)( \sqrt {2}-1)}= \dfrac { \sqrt {2}-1}{( \sqrt {2})^{2}-1}= \dfrac { \sqrt {2}-1}{1}= \sqrt {2}-1\).
              例\(2\):\( \dfrac {1}{ \sqrt {3}+ \sqrt {2}}= \sqrt {3}- \sqrt {2}\),\( \dfrac {1}{ \sqrt {4}+ \sqrt {3}}= \sqrt {4}- \sqrt {3}\),\( \dfrac {1}{ \sqrt {5}+ \sqrt {4}}= \sqrt {5}- \sqrt {4}\)
              利用以上结论解答以下问题:
              \((1) \dfrac {1}{ \sqrt {6}+ \sqrt {5}}=\) ______
              \((2)\)应用上面的结论,求下列式子的值.
              \( \dfrac {1}{ \sqrt {2}+1}+ \dfrac {1}{ \sqrt {3}+ \sqrt {2}}+ \dfrac {1}{ \sqrt {4}+ \sqrt {5}}+…+ \dfrac {1}{ \sqrt {100}+ \sqrt {99}}\)
              \((3)\)拓展提高,求下列式子的值.
              \( \dfrac {1}{1+ \sqrt {3}}+ \dfrac {1}{ \sqrt {3}+ \sqrt {5}}+ \dfrac {1}{ \sqrt {5}+ \sqrt {7}}+…+ \dfrac {1}{ \sqrt {2015}+ \sqrt {2017}}\).
            • 6.
              化简\(( \sqrt {3}-2)^{2006}⋅( \sqrt {3}+2)^{2007}\)的结果为\((\)  \()\)
              A.\(-1\)
              B.\( \sqrt {3}-2\)
              C.\( \sqrt {3}+2\)
              D.\(- \sqrt {3}-2\)
            • 7.
              计算:
              \((1)(2 \sqrt {3}+ \sqrt {6})(2 \sqrt {3}- \sqrt {6})\)
              \((2) \sqrt {48}÷ \sqrt {3}- \sqrt { \dfrac {1}{2}}× \sqrt {12}+ \sqrt {24}\).
            • 8.

              已知\(x=2- \sqrt{3} \),求代数式\((7+4 \sqrt{3}){x}^{2}+(2+ \sqrt{3})x+ \sqrt{3} \)的值.

            • 9.

               阅读理解,我们把\(\left| \begin{matrix} & a{ }b \\ & c{ }d \\ \end{matrix} \right|\)称作二阶行列式,规定它的运算法则为\(\left| \begin{matrix} & a{ }b \\ & c{ }d \\ \end{matrix} \right|=ad-bc\),例如\(\left| \begin{matrix} & {2 3} \\ & {4 5} \\ \end{matrix} \right|=2\times 5-3\times 4=-2\),请根据阅读理解解答下列各题:

              \(⑴\)计算:\(\left| \begin{matrix} & \sqrt{24}{ 3}\sqrt{6} \\ & \sqrt{\dfrac{1}{2}}{ }\sqrt{2} \\ \end{matrix} \right|\)                    

              \(⑵\)计算:\(\left| \begin{matrix} & {1 2} \\ & {3 4} \\ \end{matrix} \right|+\left| \begin{matrix} & {5 6} \\ & {7 8} \\ \end{matrix} \right|+\cdots +\left| \begin{matrix} & {97 98} \\ & {99 100} \\ \end{matrix} \right|\)          

              \(⑶\)已知实数\(a\),\(b\)满足行列式\(\left| \begin{matrix} & { }a{ }-1 \\ & -{{a}^{2}}+b{ }a-1 \\ \end{matrix} \right|=5\),则代数式\(\dfrac{{{a}^{2}}+{{b}^{2}}}{2}-ab+2\)的值  

            • 10.

              计算:

              \((1)\left( \sqrt{12}+\sqrt{1\dfrac{1}{3}} \right)\times \sqrt{3}\)

              \((2)\sqrt{48}{-}\sqrt{54}\div \sqrt{2}+\left( 3-\sqrt{3} \right)\left( 3+\sqrt{3} \right)\)

            0/40

            进入组卷