Odpowiedzi

Najlepsza Odpowiedź!
2009-11-18T11:16:15+01:00
1.
4(x+7)(x+6) - (2x+3)² + 1 = 0
4(x²+13x+42)-(4x²+12x+9)+1=0
4x²+52x+168-4x²-12x-9+1=0
40x+160=0
40x=-160
x=-40

2.
2(x-2)² ≥ (x-3)(x+3) - x + x² - 3
2(x²-4x+4)≥x²-9-x+x²-3
2x²-8x+8≥2x²-x-12
2x²-8x+8-2x²+x+12≥0
-7x+20≥0
7x≤20
x≤20/7
x≤2 i 6/7
2009-11-18T11:17:15+01:00
4(x+7)(x+6) - (2x+3)² + 1 = 0

4x+28 * x + 6 - 2x² + 2*2x*3 + 3² +1 = 0

4x+28 * x + 6 - 2x² + 12x + 9 + 1 = 0
4x + 28x + 6 - 2x² + 12x + 9 + 1 = 0
32x + 6 - 2x² + 12x + 9 + 1 = 0
44x +16 - 2x²=0

2(x-2)² ≥ (x-3)(x+3) - x + x² - 3

2(x-2)² =

2 * x² - 2*x*2 + 2² =
2 * x² - 4x + 4 =
2x² - 8x + 8


(x-3)(x+3) - x + x² - 3
Tego nie umeim
2009-11-18T11:39:25+01:00
1
4(x+7)*(x+6)-(2x+3)²+1=0
4(x²+6x+7x+42)-(4x²+12x+9)+1=0
4x²+24x+28x+168-4x²-12x-9+1=0
40x=9-1-168
40x=-160
x=-4
2
2(x-2)²≥(x-3)*(x+3)-x+x²-3
2(x²-4x+4)≥x²-9-x+x²-3
2x²-8x+8-x²+9+x-x²+3≥0
-7x≥-8-9-3
-7x≥-20
x≤20/7
x∈<-nieskończoność,20/7>