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{(y-2)² + 3x + 1 = (y+2)² -7
{x²-2x+1+2y=x²+6x+9-4y-2
{y²-4y+4+3x+1=y²+4y+4-7
{-2x+1+2y=6x+9-4y-2
{-4y+4+3x+1=4y+4-7
{-8x+6y=6 |*3
{3x-8y=-8 |*8
{-24x+18y=18
{24x-64y=-64 (dodajemy górę i dół)
-46y=-46 |:(-46)
y=1
3x-8*1=-8
3x=0 |:3
x=0
Zad.2
a) (x-√2 )(x+2√2)=x²+1
x²+2√2x-√2x-4=x²+1
√2x-4=1
√2x=5 |:√2
x=5/√2
b) 1+√3=x+2
1+√3-2=x
-1+√3=x
c) (x+2√2)√2 =√3x-1
√2x+4=√3x-1
√2x-√3x=-5
d) [ (x(x+1)) /4 ] - [ (2x²+1)/2 = (-3/4 x² ) - 0,25(x+2)
[(x²+x)/4]-[x²+½]=-¾x²-0,25x-½ |*4
x²+x-4x²+2=-3x²-x-2
x+2=-x-2
2x=-4 |:2
x=-2
a) [ (x+2)/3 ] - [ (x+4)/4 ] > (x+1)/12 |*12
4x+8-3x+12 > x+1
x+20 > x+1
b) [ ( (x- (3-2x)) /2) / 5 ] > 2 |*5
[(x-(3-2x))/2] > 10 |*2
x-3+2x > 20
3x > 23 |:3
x > 7⅔
c) [ ( ((x-2,6)/3) - 2x ) /4 ] ≥ (x+2)/5 |*4
[((x-2,6)/3)-2x)] ≥ 4x+8/5 |*15
5x-13-30x ≥ 12x+24
5x-13-30 ≥ 12x+24
-43-24 ≥ 17x
-57 ≥ 17x
Nie jest dobrze zapewne wszystko, ale może cokolwiek pomogłam...