Bài 93 trang 131 SGK giải tích 12 nâng cao, Giải phương trình:...

Bài 93. Giải phương trình:

(eqalign{
& a),{32^{{{x + 5} over {x – 7}}}} = 0,{25.128^{{{x + 17} over {x – 3}}}},; cr
& c),{4^x} – {3^{x – 0,5}} = {3^{x + 0,5}} – {2^{2x – 1}},; cr} )

(eqalign{
& b),{5^{x – 1}} = {10^x}{.2^{ – x}}{.5^{x + 1}},; cr
& d),{3^{4x + 8}} – {4.3^{2x + 5}} + 28 = 2{log _2}sqrt 2 . cr} )

Giải

a) Ta có: ({32^{{{x + 5} over {x – 7}}}} = 0,{25.128^{{{x + 17} over {x – 3}}}} Leftrightarrow {2^{{{5left( {x + 5} ight)} over {x – 7}}}} = {1 over 4}{.2^{{{7left( {x + 17} ight)} over {x – 3}}}})
( Leftrightarrow {2^{{{5left( {x + 5} ight)} over {x – 7}}}} = {2^{{{7left( {x + 17} ight)} over {x – 3}}-2}} Leftrightarrow {{5left( {x + 5} ight)} over {x – 7}} = {{7left( {x + 17} ight)} over {x – 3}} – 2,,left( 1 ight))
Điều kiện: (x e 3;,x e 7.)

(1) ( Leftrightarrow 5left( {x + 5} ight)left( {x – 3} ight) = 7left( {x + 17} ight)left( {x – 7} ight) – 2left( {x – 7} ight)left( {x – 3} ight))
( Leftrightarrow 80x = 800 Leftrightarrow x = 10) (nhận)
Vậy (S = left{ {10} ight})
(b),{5^{x – 1}} = {10^x}{.2^{ – x}}{.5^{x + 1}} Leftrightarrow {1 over 5}{.5^x} = {{{{10}^x}} over {{2^x}}}{.5.5^x} Leftrightarrow {1 over 5} = {5^x}.5 Leftrightarrow {5^x} = {1 over {25}} Leftrightarrow x =  – 2)
Vậy (S = left{ { – 2} ight})

(eqalign{
& c),,{4^x} – {3^{x – 0,5}} = {3^{x + 0,5}} – {2^{2x – 1}} Leftrightarrow {4^x} + {1 over 2}{.4^x} = {3^{x – 0,5}} + {3^{x + 0,5}} cr
& ,, Leftrightarrow {3 over 2}{.4^x} = {3^{x – 0,5}}left( {1 + 3} ight) Leftrightarrow {1 over 2}{.4^{x – 1}} = {3^{x – 1,5}} cr
& ,, Leftrightarrow {4^{x – 1,5}} = {3^{x – 1,5}} Leftrightarrow {left( {{4 over 3}} ight)^{x – 1,5}} = 1 Leftrightarrow x – 1,5 = 0 cr
& ,,, Leftrightarrow x = 1,5 cr} )

Vậy (S = left{ {1,5} ight})
d) Đặt (t = {3^{2x + 4}},left( {t > 0} ight))
Ta có phương trình: ({t^2} – 12t + 28 = 1 Leftrightarrow {t^2} – 12t + 27 = 0)

(eqalign{
& Leftrightarrow left[ matrix{
t = 9 hfill cr
t = 3 hfill cr} ight. Leftrightarrow left[ matrix{
{3^{2x + 4}} = 9 hfill cr
{3^{2x + 4}} = 3 hfill cr} ight. cr
& Leftrightarrow left[ matrix{
2x + 4 = 2 hfill cr
2x + 2 = 1 hfill cr} ight. Leftrightarrow left[ matrix{
x = – 1 hfill cr
x = – {3 over 2} hfill cr} ight. cr} )

Vậy (S = left{ { – {3 over 2}; – 1} ight})