Вправа 594 алгебра Бевз гдз 7 клас
№ 594
а) (х2 + 1)(х4 – х2 + 1) = 1
х6 + 1 = 1
х6 = 0
х = 0;
б) (х + 3)(х2 – 3х + 9) – (х – 5)(х2 + 5х + 25) = 4(20 – х)
х3 + 27 – х3 + 125 = 80 – 4х
4х = 80 – 27 – 125
4х = –72
х = –18;
в) (х – 2)(х2 + 2х + 4) – х(х + 3)(х – 3) = 1
х3 – 8 – х3 + 9х = 1
9х = 1 + 8
х = 9
х = 9 : 9
х = 1;
г) (4х2 – 1)(4х2 – 2х + 1) = 8х3(2х – 1)
(2х – 1)(2х + 1)(4х2 – 2х + 1) = 8х3(2х – 1)
(2х – 1)(8х3 + 1) = 8х3(2х – 1)
16х4 + 2х – 8х3 – 1 = 16х4 – 8х3
2х – 1 = 0
2х = 1
х = 2.
вправи поруч
