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            • 1.

              已知\((x^{2}+y^{2}+2)^{2}=9\),则\(x^{2}+y^{2}=\)_____.

            • 2.

              用换元法解方程\(2{{x}^{2}}+8x-\dfrac{20}{{{x}^{2}}+4x}=13\),若设\({{x}^{2}}+4x=y\),则原方程可化为关于\(y\)的整式方程为______________________.

            • 3.

              若\((x^{2}+y^{2})^{2}-5(x^{2}+y^{2})=6\),则\(x^{2}+y^{2}=\)____.

            • 4.

              已知\(x\)为实数,且满足\((x^{2}+3x)^{2}+2(x^{2}+3x)-3=0\),那么\(x^{2}+3x\)的值为\((\)   \()\)

              A.\(1\)             
              B.\(-3\)或\(1\)       
              C.\(3\)                
              D.\(-1\)或\(3\)
            • 5.

              若\(({{x}^{2}}+{{y}^{2}} )({{x}^{2}}+{{y}^{2}} +2)-8=0\),则\({{x}^{2}}+{{y}^{2}}\)的值为\((\)    \()\)


              A.\(-4\)或\(2\)    
              B.\(-2\)或\(4\)       
              C.\(4\)               
              D.\(2\)
            • 6.

              \((1)1\)纳米\({=}0{.}000000001\)米,则\(0{.}25\)纳米用科学记数法表示为_________   米


              \((2)\)若\(4x^{2}{+}mx{+}1\)是完全平方式,则\(m{=}\) ______ .

              \((3)\)若\(3^{m}{=}6{,}9^{n}{=}2\),则\(3^{2m{+}4n}\)的值是______ .

              \((4)\)化简  \((1-\dfrac{1}{m+1})(m+1)\)的结果是              .

              \((5)\)若\(x={{2}^{m}}+1\)\(y=3+{{4}^{m}}\)  ,则用\(x\)的代数式表示\(y\)为__________________

              \((6)\)已知\({{(a-2015)}^{2}}+{{(a-2017)}^{2}}=10\),则\({{(a-2016)}^{2}}\)的值是_____________________

            • 7.

              \(14.\)若实数\(x\)\(y\)满足\((\)\(x\)\({\,\!}^{2}+\)\(y\)\({\,\!}^{2}+2)(\)\(x\)\({\,\!}^{2}+\)\(y\)\({\,\!}^{2}-1)=0\),则\(x\)\({\,\!}^{2}+\)\(y\)\({\,\!}^{2}=\)__________

            • 8.

              已知\(x\)为实数,且满足\((x^{2}+x)^{2}+2(x^{2}+x)-3=0\),那么\(x^{2}+x\)的值为\((\)    \()\)


              A.\(1\)
              B.\(-3\)或\(1\)
              C.\(3\)
              D.\(-1\)或\(3\)
            • 9.

              若\((x^{2}+y^{2})^{2}-2(x^{2}+y^{2})-3=0\),则\(x^{2}+y^{2}=\)          .

            • 10.

              为解方程\({(}x^{2}{-}1{{)}}^{2}{-}5(x^{2}{-}1){+}4{=}0\),我们可以将\(x^{2}{-}1\)视为一个整体,然后设\(x^{2}{-}1{=}y{,}\)则\({(}x^{2}{-}1{{)}}^{2}{=}y^{2}{,}\)原方程化为\(y^{2}{-}5y{+}4{=}0{,}\)解得\(y_{1}{=}1{,}y_{2}{=}4{.}\)

              当\(y=1\)时,\(x^{2}{-}1{=}1{,}\)得\(x{=}{±}\sqrt{2}\);当\(y=4\)时,\(x^{2}{-}1{=}4{,}\)得\(x{=}{±}\sqrt{5}.\) 故原方程的解为\(x_{1}{=}\sqrt{2}{,}x_{2}{=}{-}\sqrt{2}{,}x_{3}{=}\sqrt{5}{,}x_{4}{=}{-}\sqrt{5}{.}\)在解方程的过程中,我们将\(x^{2}{-}1\)用\(y\)替换,先解出关于\(y\)的方程,达到了降低方程次数的目的,这种方法叫做“换元法”,体现了转化的数学思想。

                请根据以上的阅读,解下列方程:

              \((1)x^{4}{-}x^{2}{-}6{=}0\)      

              \((2){(}\dfrac{1}{2}x{-}1{{)}}^{2}{-}(\dfrac{1}{2}x{-}1){-}1{=}0\)

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