√(х - 1) = 6
ОДЗ: х - 1 > 0 → x > 1
возводим в квадрат
х - 1 = 36
х = 37
1) (5х+2)²=25x^2+10x+4
2) (1/3а²-1/9)²=1/9 a^4-2/27 a^2+1/81
3) (х-5у)²-(2х+у)²=(x-5y-2x-y)(x-5y+2x+y)=(-x-6y)(3x-4y)
4) (4х-3)²=16x^2-24x+9
5) (1/3а²+1/5)²=1/9 a^4+1/15a^2+1/25
<span>6) (5х+у)²-(х-2у)²=(5x+y-x+2y)(5x+y+x-2y)=(4x+3y)(6x-y)</span>
(17 + 2 √17*3 + 3) / 10 + √51 = (20 + 2 √51) / 10 + √51 = 2( 10+√51) / 10+√51 = 2
Ax^2 + 4ax + 4a = a( x^2 + 4x + 4 ) = a( x + 2 )^2
1) = 3a( b - 5 ) - 12( b - 5 ) = ( 3a - 12 )( b - 5 ) = 3( a - 4 )( b - 5 )
2) = ( a^2 )^2 - ( 25 )^2 = ( a^2 - 25 )( a^2 + 25 ) = ( a - 5 )( a + 5 )( a^2 + 25 )