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17. Efectuați calculele:
a) 6x + [2x+(3x + 5) – 3x²(2x - 4)] :(–2x);
b) 3x – [4x? (2x + 3) – 3x2(x2 + 4x)] :(-5x”);
c) x(x - 1) + [2x(x + 2)-(x2 + 4x)] :x";
d) [x(x2 + x - 2) - x(x - 2)] : x + x(x - 1).
18. Efectuați calculele:
a) (x + 2)(x2 - x + 3) - (x + 1)(x2 + 2x - 5);
b) (2x + 1)(x2 - 2x + 4) - (x - 3)(2x2 + x + 3);
c) (x – 3)(x2 – 3x + 2) - (x + 3)(x² – 2x + 3);
d) (2x - 1)(-x2 + x - 2) + (2x - 3)(x2 - x + 1).​

Răspuns :

Iamdelim

Răspuns:

17. a) 6x³+6x²-5x-5 totul supra 2x

b) -3x³-12x²+23x+12 totul supra 5

c) x²

d) 2x²-x

18. a) -2x²+4x+11

b) 2x²+6x+13

c) -7x²+14x-15

d) -2x²-1

Explicație pas cu pas:

17. a) 6x+(2x+3x+5-6x³+6x²)÷(-2x) =

= 6x+(5x+5-6x³+6x²)÷(-2x) =

= 6x+ 5x+5-6x³+6x²/-2x =

= 6x- 5x+5-6x³+6x²/2x =

= 12x²-(5x+5-6x³+6x²)/2x =

= 12x²-5x-5+6x³-6x²/2x =

= 6x²-5x-5+6x³/2x =

= 6x³+6x²-5x-5/2x

b) 3x-(8x²+12x-3x⁴-12x³)÷(-5x) =

= 3x+ 8x²+12x-3x⁴-12x³/5x =

= 3x+ x(8x+12-3x³-12x²)/5x =

= 3x+ 8x+12-3x³-12x²/5 =

= 15x+8x+12-3x³-12x²/5 =

= 23x+12-3x³-12x²/5 =

= -3x³-12x²+23x+12/5

c) x²-x+(2x²+4x-x²-4x)÷x =

= x²-x+x²÷x =

= x²-x+x =

= x²

d) (x³+x²-2x-x²+2x)÷x+x²-x =

= x³÷x+x²-x =

= x²+x²-x =

= 2x²-x

18. a) x³-x²+3x+2x²-2x+6-(x³+2x²-5x+x²+2x-5) =

= x³-x²+3x+2x²-2x+6-(x³+3x²-3x-5) =

= x³-x²+3x+2x²-2x+6-x³-3x²+3x+5 =

= -2x²+4x+11

b) 2x³-4x²+8x+x²-2x+4-(2x³+x²+3x-6x²-3x-9) =

= 2x³-4x²+8x+x²-2x+4-(2x³-5x²-9) =

= 2x³-4x²+8x+x²-2x+4-2x³+5x²+9 =

= 2x²+6x+13

c) x³-3x²+2x-3x²+9x-6-(x³-2x²+3x+3x²-6x+9) =

= x³-3x²+2x-3x²+9x-6-(x³+x²-3x+9) =

= x³-3x²+2x-3x²+9x-6-x³-x²+3x-9=

= -7x²+14x-15

d) -2x³+2x²-4x+x²-x+2+2x³-2x²+2x-3x²+3x-3 =

= 0-2x²-1 =

= -2x²-1

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