换元思想

(\sqrt{2} + \sqrt{3} + \sqrt{5})(1 + \sqrt{2} + \sqrt{3}) - (1 + \sqrt{2} + \sqrt{3} + \sqrt{5})(\sqrt{2} + \sqrt{3})
(\sqrt{2} + \sqrt{3} + \sqrt{5})(1 + \sqrt{2} + \sqrt{3}) - (1 + \sqrt{2} + \sqrt{3} + \sqrt{5})(\sqrt{2} + \sqrt{3})

若

x = \frac{1}{2}\left[\sqrt{2020} - \frac{1}{\sqrt{2020}}\right]
x = \frac{1}{2}\left[\sqrt{2020} - \frac{1}{\sqrt{2020}}\right]
,求
\sqrt{x^2 + 1} + x
\sqrt{x^2 + 1} + x
的值

化简:

\dfrac{\left[ a - \dfrac{ac^2}{(a + b)^2} \right] \left[ 1 + \dfrac{c}{a + b} + \dfrac{c^2}{(a + b)^2} \right]}{\left( b + \dfrac{bc}{a + b} \right) \left[ 1 - \dfrac{c^3}{(a + b)^3} \right]}
\dfrac{\left[ a - \dfrac{ac^2}{(a + b)^2} \right] \left[ 1 + \dfrac{c}{a + b} + \dfrac{c^2}{(a + b)^2} \right]}{\left( b + \dfrac{bc}{a + b} \right) \left[ 1 - \dfrac{c^3}{(a + b)^3} \right]}

如果

\dfrac{b}{a} + \dfrac{a}{b} = 4
\dfrac{b}{a} + \dfrac{a}{b} = 4
,那么
\dfrac{(a + b)^2}{(a - b)^2}
\dfrac{(a + b)^2}{(a - b)^2}
的值为( )

执果索因

1.(1)

规定两数

 m, n
m, n
的运算
[m, n]:若  m^k = n
[m, n]:若 m^k = n
,则
[m, n] = k
[m, n] = k
(如
[2, 8] = 3
[2, 8] = 3
,因
 2^3 = 8
2^3 = 8
)。已知
 [a, 16] + [a, 9] = [a, t]
[a, 16] + [a, 9] = [a, t]
(
a \gt 0  且  a \neq 1
a \gt 0 且 a \neq 1
),求
 t
t
的值。

(2)

小明用积的乘方巧算:

(pq)^n = p^n q^n
(pq)^n = p^n q^n
。若
 x = 3^2
x = 3^2
,
y = 2^4
y = 2^4
,请通过拆幂、换元,用
 x 、 y
x 、 y
表示 6^8 。

2.已知

  a \neq b
a \neq b
,且
a + \frac{1}{b} = b + \frac{1}{a}
a + \frac{1}{b} = b + \frac{1}{a}
,求 ab 的值;


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