вправа 2.55 гдз 11 клас алгебра Істер 2019
Вправа 2.55
Умова:
Розв'яжіть рівняння:
1) (2 - √3)х + (2 + √3)х = 4;
2) 4(√5 - 2)х-10 = (2/√5 + 2)х-10;
3) (2√2 + 3)х + (2√2 - 3)х = 34.
1) (2 - √3)х + (2 + √3)х = 4;
2) 4(√5 - 2)х-10 = (2/√5 + 2)х-10;
3) (2√2 + 3)х + (2√2 - 3)х = 34.
Відповідь:
1) (2 - √3)х + (2 + √3)х = 4
(2 - √3)(2 + √3) = 4 - 3 = 1 =>
2 - √3 = 1/(2 + √3)
нехай: (2 - √3)х = t, t > 0
t + 1/t - 4 = 0
t2 - 4t + 1 = 0
Д = (-4)2 - 4 = 12
t1;2 = (4±√12)/2
t1;2 = (2(2±√3)/2
t1 = 2 - √3, t2 = 2 + √3
(2 - √3)х = 2 - √3 (2 - √3)х = 2 + √3
х = 1 х = -1;
(2 - √3)(2 + √3) = 4 - 3 = 1 =>
2 - √3 = 1/(2 + √3)
нехай: (2 - √3)х = t, t > 0
t + 1/t - 4 = 0
t2 - 4t + 1 = 0
Д = (-4)2 - 4 = 12
t1;2 = (4±√12)/2
t1;2 = (2(2±√3)/2
t1 = 2 - √3, t2 = 2 + √3
(2 - √3)х = 2 - √3 (2 - √3)х = 2 + √3
х = 1 х = -1;
2) 4(√5 - 2)х-10 = (2/√5 + 2)х-10
(2х-10)/((√5 + 2)х-10 • (√5 - 2)х-10 • 4) = 1
(2х-10)/(4((√5 + 2)(√5 - 2)х-10)) = 1
(2х-10)/(4 • (5 - 4)х-10) = 1
(2х-10)/(4 • 1х-10) = 1
2х-10 = 4
2х-10 = 22
х - 10 = 2
х = 12;
(2х-10)/((√5 + 2)х-10 • (√5 - 2)х-10 • 4) = 1
(2х-10)/(4((√5 + 2)(√5 - 2)х-10)) = 1
(2х-10)/(4 • (5 - 4)х-10) = 1
(2х-10)/(4 • 1х-10) = 1
2х-10 = 4
2х-10 = 22
х - 10 = 2
х = 12;
3) (2√2 + 3)х + (2√2 - 3)х = 34
(2√2 + 3) • (2√2 - 3) = 8 - 9 = -1
(2√2 + 3) = -1/(3 - 2√2)
-(3 - 2√2) = 1/(2√2 + 3)
нехай: -(3 - 2√2)х = t, t > 0
-t + 1/t - 34 = 0
-t2 - 34t - 1 = 0
t2 + 34t + 1 = 0
Д = (-34)2 - 4 • (-1) • (-1) = 1152
√1152 = 24√2
t1;2 = (-34±24√2)/-2 = (-2(17±12√2))/-2
t1 = 17 - 12√2, t2 = 17 + 12√2
(-3 - 2√2)х = 32 - 12√2 + (√8)2
(-3 2√2)х = (3 - 2√2)2
х = 2
(3 - 2√2)х = 17 + 12√2
(3 - 2√2)х = (3 + 2√2)-2
х = -2.
(2√2 + 3) • (2√2 - 3) = 8 - 9 = -1
(2√2 + 3) = -1/(3 - 2√2)
-(3 - 2√2) = 1/(2√2 + 3)
нехай: -(3 - 2√2)х = t, t > 0
-t + 1/t - 34 = 0
-t2 - 34t - 1 = 0
t2 + 34t + 1 = 0
Д = (-34)2 - 4 • (-1) • (-1) = 1152
√1152 = 24√2
t1;2 = (-34±24√2)/-2 = (-2(17±12√2))/-2
t1 = 17 - 12√2, t2 = 17 + 12√2
(-3 - 2√2)х = 32 - 12√2 + (√8)2
(-3 2√2)х = (3 - 2√2)2
х = 2
(3 - 2√2)х = 17 + 12√2
(3 - 2√2)х = (3 + 2√2)-2
х = -2.