Bài 8 trang 78 SGK Đại số và Giải tích 12 Nâng cao, Đơn giản biểu thức:...

Bài 8. Đơn giản biểu thức:

a) ({{sqrt a  – sqrt b } over { oot 4 of a  – oot 4 of b }} – {{sqrt a  + oot 4 of {ab} } over { oot 4 of a  + oot 4 of b }});                     

b) ({{a – b} over { oot 3 of a  – oot 3 of b }} – {{a + b} over { oot 3 of a  + oot 3 of b }});

c) (left( {{{a + b} over { oot 3 of a  + oot 3 of b }} – oot 3 of {ab} } ight):{left( { oot 3 of a  – oot 3 of b } ight)^2};) 

d) ({{a – 1} over {{a^{{3 over 4}}} + {a^{{1 over 2}}}}}.{{sqrt a  + oot 4 of a } over {sqrt a  + 1}}.{a^{{1 over 4}}} + 1.)

Giải

a) ({{sqrt a  – sqrt b } over { oot 4 of a  – oot 4 of b }} – {{sqrt a  + oot 4 of {ab} } over { oot 4 of a  + oot 4 of b }} = {{left( { oot 4 of a  + oot 4 of b } ight)left( { oot 4 of a  – oot 4 of b } ight)} over { oot 4 of a  – oot 4 of b }} – {{ oot 4 of a left( { oot 4 of a  + oot 4 of b } ight)} over { oot 4 of a  + oot 4 of b }})

( = oot 4 of a  + oot 4 of b  – oot 4 of a  = oot 4 of b )

b) ({{a – b} over { oot 3 of a  – oot 3 of b }} – {{a + b} over { oot 3 of a  + oot 3 of b }} = {{{{left( { oot 3 of a } ight)}^3} – {{left( { oot 3 of b } ight)}^3}} over { oot 3 of a  – oot 3 of b }} – {{{{left( { oot 3 of a } ight)}^3} + {{left( { oot 3 of b } ight)}^3}} over { oot 3 of a  + oot 3 of b }})

( = oot 3 of {{a^2}}  + oot 3 of {ab}  + oot 3 of {{b^2}}  – left( { oot 3 of {{a^2}}  – oot 3 of {ab}  + oot 3 of {{b^2}} } ight) = 2 oot 3 of {ab} )

c) (left( {{{a + b} over { oot 3 of a  + oot 3 of b }} – oot 3 of {ab} } ight):{left( { oot 3 of a  – oot 3 of b } ight)^2} = left( { oot 3 of {{a^2}}  – oot 3 of {ab}  + oot 3 of {{b^2}}  – oot 3 of {ab} } ight):{left( { oot 3 of a  – oot 3 of b } ight)^2})

( = left( { oot 3 of {{a^2}}  – 2 oot 3 of {ab}  + oot 3 of {{b^2}} } ight):{left( { oot 3 of a  – oot 3 of b } ight)^2} = {left( { oot 3 of a  – oot 3 of b } ight)^2}:{left( { oot 3 of a  – oot 3 of b } ight)^2} = 1)

d) ({{a – 1} over {{a^{{3 over 4}}} + {a^{{1 over 2}}}}}.{{sqrt a  + oot 4 of a } over {sqrt a  + 1}}.{a^{{1 over 4}}} + 1. = {{left( {sqrt a  + 1} ight)left( {sqrt a  – 1} ight)} over {sqrt a left( { oot 4 of a  + 1} ight)}}.{{ oot 4 of a left( { oot 4 of a  + 1} ight)} over {left( {sqrt a  + 1} ight)}}. oot 4 of a  + 1)

                                        ( = sqrt a  – 1 + 1 = sqrt a ).