Bài 85 trang 130 SGK giải tích 12 nâng cao, Cho x < 0. Chứng minh rằng:...
Cho x Bài 85 . Cho (x < 0). Chứng minh rằng: (sqrt {{{ – 1 + sqrt {1 + {1 over 4}{{left( {{2^x} – {2^{ – x}}} ight)}^2}} } over {1 + sqrt {1 + {1 over 4}{{left( {{2^x} – {2^{ – x}}} ight)}^2}} }}} = {{1 – {2^x}} over {1 + {2^x}}}) Giải Ta có: (1 + {1 over 4}{left( {{2^x} – {2^{ – ...
Bài 85. Cho (x < 0). Chứng minh rằng: (sqrt {{{ – 1 + sqrt {1 + {1 over 4}{{left( {{2^x} – {2^{ – x}}} ight)}^2}} } over {1 + sqrt {1 + {1 over 4}{{left( {{2^x} – {2^{ – x}}} ight)}^2}} }}} = {{1 – {2^x}} over {1 + {2^x}}})
Giải
Ta có: (1 + {1 over 4}{left( {{2^x} – {2^{ – x}}} ight)^2} = {1 over 4}left( {4 + {4^x} – 2 + {4^{ – x}}} ight) = {1 over 4}left( {{4^x} + 2 + {4^{ – x}}} ight) = {1 over 4}{left( {{2^x} + {2^{ – x}}} ight)^2})
Do đó:
(eqalign{
& sqrt {{{ – 1 + sqrt {1 + {1 over 4}{{left( {{2^x} – {2^{ – x}}}
ight)}^2}} } over {1 + sqrt {1 + {1 over 4}{{left( {{2^x} – {2^{ – x}}}
ight)}^2}} }}} = sqrt {{{ – 1 + {1 over 2}left( {{2^x} + {2^{ – x}}}
ight)} over {1 + {1 over 2}left( {{2^x} + {2^{ – x}}}
ight)}}} = sqrt {{{{2^x} – 2 + {2^{ – x}}} over {{2^x} + 2 + {2^{ – x}}}}} cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, = sqrt {{{{2^x} – 2 + {1 over {{2^x}}}} over {{2^x} + 2 + {1 over {{2^x}}}}}} = sqrt {{{{4^x} – {{2.2}^x} + 1} over {{4^x} + {{2.2}^x} + 1}}} = sqrt {{{{{left( {{2^x} – 1}
ight)}^2}} over {{{left( {{2^x} + 1}
ight)}^2}}}} = {{1 – {2^x}} over {1 + {2^x}}} cr} )
(vì với (x < 0) thì ({2^x} < 1))
