вправа 6.35 гдз 11 клас алгебра Істер 2019
Вправа 6.35
Розв'яжіть рівняння:
1) (3log2х-10)/(log2х+4) = (log2х-4)/(log2х+1);
2) (log3х)/(2log3х-6) + 9/(9-log32х) = 8/(2log3х+6);
3) 2lg2х + (1 - √2)lgх2 = 2√2;
4) 4 - log2х = 3√log2х.
1) (3log2х-10)/(log2х+4) = (log2х-4)/(log2х+1);
2) (log3х)/(2log3х-6) + 9/(9-log32х) = 8/(2log3х+6);
3) 2lg2х + (1 - √2)lgх2 = 2√2;
4) 4 - log2х = 3√log2х.
Умова:
Відповідь:
1) (3log2х-10)/(log2х+4) = (log2х-4)/(log2х+1)
(3log2х - 10)(log2х + 1) = (log2х + 4)(log2х - 4)
3log22х + 3log2х - 10log2х - 10 = log22х - 4log2х + 4log2х - 16
3log22х - log22х - 7log2х - 10 + 16 = 0
2log22х - 7log2х + 6 = 0
ОДЗ: х > 0
заміна: log2х = t
2t2 - 7t + 6 = 0
Д = (-7)2 - 4 • 2 • 6 = 49 - 48 = 1
t1;2 = (7±1)/4 = 2; 1,5
log2х = t1 log2х = t2
log2х = 2 log2х = 1,5
х = 22 х = 23/2
х1 = 4 х = √23 = √8
х2 = 2√2;
2) (log3х)/(2log3х-6) + 9/(9-log32х) = 8/(2log3х+6)
(log3х)/(2(log3х-3)) - 9/(log32х-9) = 8/(2(log3х+3))
(log3х)/(2(log3х-3)) - 9/(log32х-9) - 8/(2(log3х+3)) = 0
log32х + 3log3х - 18 - 8log3х + 24 = 0
ОДЗ: х > 0
log32х - 5log3х + 6 = 0
заміна: log3х = t
t2 - 5t + 6 = 0
Д = (-5)2 - 4 • 6 = 1
t1;2 = (5±1)/2 = 3; 2
log3х = t1 log3х = t2
log3х = 3 log3х = 2
х = 33 х = 32
х = 27 х = 9;
3) 2lg2х + (1 - √2)lgх2 = 2√2
2lg2х + 2(1 - √2)lgх - 2√2 = 0 : 2
lg2х + (1 - √2)lgх - √2 = 0
ОДЗ: х > 0
заміна: lgх = t
t2 + (1 - √2)t - √2 = 0
Д = (1 - √2)2 + 4 • √2 = 1 - 2√2 + (√2)2 + 4√2 = 1 - 2√2 +
2 + 4√2 = 3 + 2√2 = 1 + 2√2 + 2 = (1 + √2)2
√Д = √(1+√2)2 = 1 + √2
t1;2 = (-(1-√2)±(1+√2))/2
t1 = (-1+√2+1+√2)/2 = (2√2)/2 = √2
t2 = (-1+√2-1-√2)/2 = -2/2 = -1
lgх = t1 lgх = t2
lgх = √2 lgх = -1
х = 10√2 х = 0,1;
4) 4 - log2х = 3√log2х
ОДЗ: х > 0
заміна: √log2х = t
4 - t2 = 3t
-t2 - 3t + 4 = 0
Д = (-3)2 - 4 • 4 • (-1) = 25
t1;2 = (3±5)/-2 = -4; 1
√log2х = t1 √log2х = t2
√log2х = -4 √log2х = 1
Ø log2х = 1
х = 2.
(3log2х - 10)(log2х + 1) = (log2х + 4)(log2х - 4)
3log22х + 3log2х - 10log2х - 10 = log22х - 4log2х + 4log2х - 16
3log22х - log22х - 7log2х - 10 + 16 = 0
2log22х - 7log2х + 6 = 0
ОДЗ: х > 0
заміна: log2х = t
2t2 - 7t + 6 = 0
Д = (-7)2 - 4 • 2 • 6 = 49 - 48 = 1
t1;2 = (7±1)/4 = 2; 1,5
log2х = t1 log2х = t2
log2х = 2 log2х = 1,5
х = 22 х = 23/2
х1 = 4 х = √23 = √8
х2 = 2√2;
2) (log3х)/(2log3х-6) + 9/(9-log32х) = 8/(2log3х+6)
(log3х)/(2(log3х-3)) - 9/(log32х-9) = 8/(2(log3х+3))
(log3х)/(2(log3х-3)) - 9/(log32х-9) - 8/(2(log3х+3)) = 0
log32х + 3log3х - 18 - 8log3х + 24 = 0
ОДЗ: х > 0
log32х - 5log3х + 6 = 0
заміна: log3х = t
t2 - 5t + 6 = 0
Д = (-5)2 - 4 • 6 = 1
t1;2 = (5±1)/2 = 3; 2
log3х = t1 log3х = t2
log3х = 3 log3х = 2
х = 33 х = 32
х = 27 х = 9;
3) 2lg2х + (1 - √2)lgх2 = 2√2
2lg2х + 2(1 - √2)lgх - 2√2 = 0 : 2
lg2х + (1 - √2)lgх - √2 = 0
ОДЗ: х > 0
заміна: lgх = t
t2 + (1 - √2)t - √2 = 0
Д = (1 - √2)2 + 4 • √2 = 1 - 2√2 + (√2)2 + 4√2 = 1 - 2√2 +
2 + 4√2 = 3 + 2√2 = 1 + 2√2 + 2 = (1 + √2)2
√Д = √(1+√2)2 = 1 + √2
t1;2 = (-(1-√2)±(1+√2))/2
t1 = (-1+√2+1+√2)/2 = (2√2)/2 = √2
t2 = (-1+√2-1-√2)/2 = -2/2 = -1
lgх = t1 lgх = t2
lgх = √2 lgх = -1
х = 10√2 х = 0,1;
4) 4 - log2х = 3√log2х
ОДЗ: х > 0
заміна: √log2х = t
4 - t2 = 3t
-t2 - 3t + 4 = 0
Д = (-3)2 - 4 • 4 • (-1) = 25
t1;2 = (3±5)/-2 = -4; 1
√log2х = t1 √log2х = t2
√log2х = -4 √log2х = 1
Ø log2х = 1
х = 2.