优优班--学霸训练营 > 知识点挑题
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            • 1.

              计算:

              \((1)(a+2)^{2}+a(a-4)\);              \((2)\dfrac{{5}a-3b}{{{a}^{{}}}-{{b}^{{}}}}-\dfrac{2a}{{{a}^{{}}}-{{b}^{{}}}}\).

            • 2. 计算:
              \((1)2^{-2}+ \sqrt {8}- \dfrac {1}{2}\sin 30^{\circ}\); 
              \((2)(1+ \dfrac {1}{x-1})÷ \dfrac {x}{x^{2}-1}\).
            • 3.

              化简:\(( \dfrac{2x-1}{x+1}-x+1) ÷ \dfrac{x-2}{x^{2}+2x+1}\)

            • 4.

              化简:\(\dfrac{{{a}^{2}}-1}{{{a}^{2}}}÷\left( \dfrac{1}{a}-1 \right)·a=\)______________.

            • 5.

              \((1)\)计算:\(|-4|-\sqrt{9}-{{(2010-{ }\!\!\pi\!\!{ })}^{0}}+3\tan 30{}^\circ \)  \((2)\)化简:\(\left( a-\dfrac{1}{a} \right)\div \dfrac{{{a}^{2}}-2a+1}{a}\).

            • 6.

              对\(x\),\(y\)定义一种新运算\(T\),规定:\(T(x,y)= \dfrac{ax+by}{2x+y} (\)其中\(a\),\(b\)均为非零常数\()\),这里等式右边是通常的四则运算,例如:\(T(0,1)= \dfrac{a×0+b×1}{2×0+1}=b \).

                  \((1)\)已知\(T(1,-1)=-2\),\(T(4,2)=1\).

                  \(①\)求\(a\),\(b\)的值.

                  \(②\)若关于\(m\)的不等式组\(\begin{cases}T(2m,5-4m)\leqslant 4, \\ T(m,3-2m) > p\end{cases} \)恰好有\(3\)个整数解,求实数\(p\)的取值范围.

                  \((2)\)若\(T(x,y)=T(y,x)\)对任意实数\(x\),\(y\)都成立\([\)这里\(T(x,y)\)和\(T(y,x)\)均有意义\(]\),则\(a\),\(b\)应满足怎样的关系式?

            • 7.

              化简:\( \dfrac{a^{2}-ab}{a^{2}}÷( \dfrac{a}{b}- \dfrac{b}{a}).\)

            • 8.

              计算:\(\dfrac{x-y}{x}\div (x-\dfrac{2xy-{{y}^{2}}}{x})=\_\_\_\_\_\_\_\_\).

            • 9.
              化简\((1+ \dfrac {1}{x-2})÷ \dfrac {x-1}{x^{2}-4x+4}\)的结果是\((\)  \()\)
              A.\(x+2\)
              B.\(x-1\)
              C.\( \dfrac {1}{x+2}\)
              D.\(x-2\)
            • 10.
              下列运算正确的是\((\)  \()\)
              A.\( \sqrt {54}⋅ \sqrt { \dfrac {1}{2}}= \dfrac {3}{2} \sqrt {6}\)
              B.\( \sqrt {(a^{3})^{2}}=a^{3}\)
              C.\(( \dfrac {1}{a}+ \dfrac {1}{b})^{2}÷( \dfrac {1}{a^{2}}- \dfrac {1}{b^{2}})= \dfrac {b+a}{b-a}\)
              D.\((-a)^{9}÷a^{3}=(-a)^{6}\)
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