1、化简: $$sqrt[a]{sqrt[b]x over sqrt[c]{x}} cdot sqrt[b]{sqrt[c]{x} over sqrt[a]{x}} cdot sqrt[c]{sqrt[a]{x} over sqrt[b]{x}}.$$ 解答: $$ ext{原式} = left(x^{{1over b} - {1over c}} ight)^{1over a} cdot left(x^{{1over c} - {1over a}} ight)^{1over b}cdotleft(x^{{1over a} - {1over b}} ight)^{1over c}$$ $$= x^{{1over ab} - {1over ac}}cdot x^{{1over bc} - {1over ab}} cdot x^{{1over ac} - {1over bc}}= x^0 = 1.$$
2、化简: $${left(sqrt x - sqrt y ight)^3 + {2x^2 over sqrt x} + ysqrt y over xsqrt x + ysqrt y} + {3sqrt{xy} - 3y over x - y}.$$ 解答: $$ ext{原式} ={left(x^{1over2} - y^{1over2} ight)^3 + 2x^{3over2} + y^{3over2} over x^{3over2} + y^{3over2}} + {3x^{1over2}cdot y^{1over2} - 3y over x - y}$$ $$= {3x^{3over2} - 3xy^{1over2} + 3x^{1over2}y over x^{3over2} + y^{3over2}} + {3y^{1over2}cdot left(x^{1over2} - y^{1over2} ight) over left(x^{1over2} + y^{1over2} ight) left(x^{1over2} - y^{1over2} ight)}$$ $$= {3x^{1over2} cdot left(x - x^{1over2}y^{1over2} + y ight) over left(x^{1over2} + y^{1over2} ight) left(x - x^{1over2}y^{1over2} + y ight)} + {3y^{1over2}cdot left(x^{1over2} - y^{1over2} ight) over left(x^{1over2} + y^{1over2} ight) left(x^{1over2} - y^{1over2} ight)}$$ $$= {3x^{1over2} + 3y^{1over2} over x^{1over2} + y^{1over2}}= 3.$$
3、计算: $${left(9 + 4sqrt5 ight)^{3over2} + left(9 - 4sqrt5 ight)^{3over2} over left(11 + 2sqrt{30} ight)^{3over2} - left(11 - 2sqrt{30} ight)^{3over2}}.$$ 解答: $$ ext{原式} = {left(sqrt5 + 2 ight)^3 + left(sqrt5 - 2 ight)^3 over left(sqrt6 + sqrt5 ight)^3 - left(sqrt6 - sqrt5 ight)^3}$$ $$= {2sqrt5cdotleft(18 - 1 ight) over 2sqrt5cdotleft(22+1 ight)}= {17 over 23}.$$
4、化简: $${1over2}lgleft(2x + 2sqrt{x^2 - 1} ight) + lgleft(sqrt{x + 1} - sqrt{x - 1} ight).$$ 解答: $$ ext{原式} = lgsqrt{2x + 2sqrt{x^2 - 1}} + lgleft(sqrt{x + 1} - sqrt{x - 1} ight)$$ $$= lgleft(sqrt{x + 1} + sqrt{x - 1} ight) + lgleft(sqrt{x + 1} - sqrt{x - 1} ight)$$ $$= lgleft(x + 1 - x + 1 ight)= lg2.$$
5、已知 $log_310 = a$, $log_625 = b$, 求 $log_445$.
解答:
由 $$log_{4}45 = {lg45 over lg 4} = {2lg3 +lg5 over 2lg2},$$ 考虑分别求出 $lg3$, $lg5$, $lg2$. $$log_310 = {lg10 over lg 3} = aRightarrowlg3 = {1over a}.$$ $$log_625 = {lg25 over lg6} = {2lg5 over lg2 + lg3} = b$$ $$Rightarrow egin{cases}2lg5 - blg2 = {bover a}\ lg5 + lg2 = 1 end{cases}Rightarrow egin{cases}lg2 = {2a - b over ab + 2a}\ lg5 = {ab + b over ab + 2a} end{cases}.$$ 因此$$log_445 = {{2over a} + {ab + b over ab + 2a} over {4a - 2b over ab + 2a}} = {ab + 3b + 4 over 2(2a - b)}.$$
6、如果 $log_8a + log_4b^2 = 5$, 且 $log_8b + log_4a^2 = 7$, 求 $ab$.
解答: $$ecause egin{cases}log_8a + log_4b^2 = 5\ log_8b + log_4a^2 = 7 end{cases}$$ $$Rightarrow log_8ab + log_4a^2b^2 = 12$$ $$Rightarrow log_8ab + log_2ab = 12$$ $$Rightarrow {1over3}log_2ab + log_2ab = 12$$ $$Rightarrow log_2ab = 9$$ $$ herefore ab = 2^9 = 512.$$
7、若 $(6.2)^x = (0.0062)^y = 1000$, 则 $displaystyle{1over x} - {1over y} = 1$.
解答: $$ecause egin{cases}x = log_{6.2}1000\ y = log_{0.0062}1000 end{cases}$$ $$Rightarrow egin{cases}{1over x} = log_{1000}6.2 = {1over3}lg6.2\ {1over y} = log_{1000}0.0062 = {1over3}lg0.0062 end{cases}$$ $$ herefore {1over x} - {1over y} = {1over3}lg{6.2over 0.0062} = 1.$$
另解: $$ecause egin{cases}6.2 = 1000^{1over x}\ 0.0062 = 1000^{1over y}end{cases}$$ $$Rightarrow1000 = 1000^{{1over x} - {1over y}}$$ $$ herefore {1over x} - {1over y} = 1.$$
8、设 $a, b$ 为不等于1的正有理数, 且方程 $x^2 - xlog_ba + displaystyle{aover b} = 0$ 的两根为 $log_ab, log_ab^3$, 求 $a, b$ 的值.
解答: $$ecause egin{cases}log_ab + log_ab^3 = log_ba\ log_ab cdot log_ab^3 = {aover b} end{cases}$$ $$Rightarrow egin{cases}4log_ab = log_ba\ 3log_a^2b = {aover b}end{cases}$$ 令 $log_ab= m$. $$Rightarrow 4m = {1over m}Rightarrow m = pm{1over 2}$$ $$Rightarrow {a over b} = {3over4}$$ $$Rightarrow log_ab = log_a{4over3}a = pm{1over2}$$ $$Rightarrow a^{1over2} = {4over3}a ext{或} a^{-{1over2}} = {4over3}a$$ $$Rightarrow a = {16over9}a^2 ext{或} {1over a} = {16over9}a^2$$ $$Rightarrow egin{cases}a_1 = {9over16}\ b_1 = {3over4} end{cases}, egin{cases}a_2 = {1over4}sqrt[3]{36}\ b_2 = {1over3}sqrt[3]{36} end{cases}.$$
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