нужно решить: 2, 3, 4заранее спасибо!. Created by dredsi. algebra-ru

нужно решить: 2, 3, 4заранее спасибо!...

dariakuz2015

Ответ:

2)4x²-12x+9+3x+15-4x²-20x=82

-29x + 24 = 82

-29x = 58

x = -2

3) x(4x-x²-12+3x)=16-x(x²-7x+12.25)

4x²-x³-12x+3x²=16-x³+x²-12.25x

-12x+12.25x=16

0.25x=16

x=64

4) 16x²-8x+1-(12x²+10x-18x-15)=4(x²-4x+4)+16x

16x²-8x+1-12x²-10x+18x+15=4x²-16x+16+16x

16=16

x - любое число

0 votes Thanks 1

Tes2able

Номер 2

$(2x - 3) {}^{2} + (3 - 4x)(x + 5) = 82 \\ 4x {}^{2} - 12x + 9 + 3x + 15 - 4x {}^{2} - 20x = 82 \\ - 29x + 24 = 82 \\ - 29x = 58 \: | \div (- 29) \\ x = - 2$

Номер 3

$x(x - 3)(4 - x) = 16 - x(x - 3.5) {}^{2} \\ (x {}^{2} - 3x)(4 - x) = 16 - x(x - 3.5) {}^{2} \\ 4x {}^{2} - {x}^{3} - 12x + 3x {}^{2} = 16 - x(x {}^{2} - 7x + 12.25) \\ 4x {}^{2} - {x}^{3} - 12x + 3x {}^{2} = 16 - x {}^{3} + 7x {}^{2} - 12.25x \\ 7x {}^{2} - 12x = 16 + 7x {}^{2} - 12.25x \\ - 12x = 16 - 12.25x \\ 0.25 = 16 \: | \div 0.25\\ x = 64$

Номер 4

$(4x - 1) {}^{2} - (2x - 3)(6 x + 5) = 4(x - 2) {}^{2} + 16x \\ 16x {}^{2} - 8x + 1 - (12x {}^{2} + 10x - 18x - 15) = 4(x {}^{2} - 4x + 4) + 16x \\ 16x {}^{2} - 8x + 1 - 12x {}^{2} - 10x + 18x + 15 = 4x {}^{2} - 16x + 16 + 16x \\ 4x {}^{2} + 16 = 4x {}^{2} + 16 \:$

х - любое число

2 votes Thanks 1

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