优优班--学霸训练营 > 知识点挑题
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            • 1.
              计算:
              \((1)(\sqrt{2\sqrt{2}})^{\frac{4}{3}}{-}4{×}(\dfrac{16}{49})^{{-}\frac{1}{2}}{-}\sqrt[4]{2}{×}8^{0{.}25}{+}({-}2014)^{0}\);
              \((2)\log_{2{.}5}6{.}25{+}\lg\dfrac{1}{100}{+}\ln(e\sqrt{e}){+}\log_{2}(\log_{2}16)\).
            • 2.
              计算\((\lg \dfrac {1}{4}-\lg 25)÷100^{- \frac {1}{2}}=\) ______ .
            • 3.
              若正数\(a\),\(b\)满足\(\log _{2}a=\log _{5}b=\lg (a+b)\),则\( \dfrac {1}{a}+ \dfrac {1}{b}\)的值为 ______
            • 4.
              \((1)\)求值:\(2\log _{3}2-\log _{3} \dfrac {32}{9}+\log _{3}8\);
              \((2)\)求函数\(f(x)= \dfrac {1}{ \sqrt {12-x}}+\log _{(x-3)}(x^{2}-x-30)\)的定义域.
            • 5.
              计算下列各式:
              \((1)(2 \dfrac {1}{4})^{ \frac {1}{2}}-(-9.6)^{0}-(3 \dfrac {3}{8})^{- \frac {2}{3}}+(1.5)^{-2}\);
              \((2)\log _{3} \dfrac { \sqrt[4]{27}}{3}+\lg 25+\lg 4+7^{\log _{7}2}\).
            • 6.
              求值:
              \((I)(2 \dfrac {1}{4})^{ \frac {1}{2}}-(-9.6)^{0}-(3 \dfrac {3}{8})^{- \frac {2}{3}}+(1.5)^{-2}\);
              \((II)\) \(\lg 14-2\lg \dfrac {7}{3}+\lg 7-\lg 18\).
            • 7.
              \((1)\)在\(\triangle ABC\)中,若\(2\lg \tan B=\lg \tan A+\lg \tan C\),则\(B\)的取值范围是 ______ .
              \((2)\)求函数\(y=7-4\sin x\cos x+4\cos ^{2}x-4\cos ^{4}x\)的最大值 ______ .
            • 8.
              \((1)\)不用计算器计算:\(2\log _{3}2-\log _{3} \dfrac {32}{9}+\log _{3}8-5^{\log _{5}3}\)
              \((2)\)如果\(f(x- \dfrac {1}{x})=(x+ \dfrac {1}{x})^{2}\),求\(f(x+1)\).
            • 9.
              \((1)\)已知\(\tan α= \dfrac {1}{3}\),求\( \dfrac {\sin α+3\cos α}{\sin \alpha -\cos \alpha }\)的值.
              \((2)\log _{3} \sqrt {27}+\lg 25+\lg 4+7^{\log \;_{7}2}+(-9.8)^{0}\).
            • 10.

              计算下列各式的值:

              \((1)2{{\log }_{3}}2-{{\log }_{3}}\dfrac{32}{9}+{{\log }_{3}}8-{{25}^{{{\log }_{5}}3}}\).

              \((2){{[{{({{0.064}^{\frac{1}{5}}})}^{-2.5}}]}^{\frac{2}{3}}}-\sqrt[3]{3\dfrac{3}{8}}-{{\mathrm{ }\!\!\pi\!\!{ }}^{0}}\).

            0/40

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