Question

1) 1 (2x+1)(2x- 1) + 20x;Б) у (у+6) < (y+3)2 ;B) (x +5)(x - 8) > x(x + 7) - 32;

Answer 1

1)4x^2 + 20 х = 0

х ( 4 х+20) =0

х=0 или x=-5

2) у^ 2 + 6 у <2 у +6

y^2+6y-2y-6<0

y^2+4y<6

y(y+4)<6

y<6 y<2

3)x^2-8x+5x-40>x^2+7x-32

-8x+5x-40-7x+32>0

-10x-12>0

-10x>12 |÷-10

x<-1,2

Answer 2

Let's solve each equation one by one:

a) 1(2x+1)(2x-1) + 20x

Expanding the expression: 1(4x^2 - 1) + 20x

Now simplify by distributing the 1: 4x^2 - 1 + 20x

Combining like terms: 4x^2 + 20x - 1

b) y(y + 6) < (y + 3)^2

Expanding the expression on the right side: y(y + 6) < (y^2 + 6y + 6y + 9)

Simplifying the right side: y^2 + 6y < y^2 + 12y + 9

Subtracting y^2 from both sides: 6y < 12y + 9

Subtracting 12y from both sides: -6y < 9

Dividing both sides by -6 (remember to reverse the inequality when dividing by a negative number): y > -9/6 or y > -3/2

c) (x + 5)(x - 8) > x(x + 7) - 32

Expanding both sides: x^2 - 3x - 40 > x^2 + 7x - 32

Canceling out the common terms on both sides (x^2): -3x - 40 > 7x - 32

Bringing all the x terms to one side: -3x - 7x > -32 + 40

Simplifying: -10x > 8

Dividing both sides by -10 (remember to reverse the inequality when dividing by a negative number): x < -8/10 or x < -4/5

So the solutions are: a) 4x^2 + 20x - 1 b) y > -3/2 c) x < -4/5