mi servono i numeri 119 120 e 121 grazie mille in anticipo
@fagottino - La 120 non si legge bene, ripostala a parte. Comunque posta una domanda per volta, come da regolamento, oltretutto avrai anche più possibili risposte. Saluti.
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$\small \left(-x+\dfrac{1}{2}x\right)\left(-4x^2+\dfrac{3}{2}x^2\right)-\left(\dfrac{2}{9}x\right)\left(-3x^2\right)-\left[\left(x^2+\dfrac{x^2}{4}\right)\left(8x\right)+\left(-\dfrac{1}{2}x^3\right)\right]-x^3 =$
$\small = \left(\dfrac{-2+1}{2}x\right)\left(\dfrac{-8+3}{2}x^2\right)-\left(-\dfrac{\cancel6^2}{\cancel9_3}x^3\right)-\left[\left(\dfrac{4+1}{4}x^2\right)\left(8x\right)-\dfrac{1}{2}x^3\right]-x^3 =$
$\small = \left(-\dfrac{1}{2}x\right)\left(-\dfrac{5}{2}x^2\right)-\left(-\dfrac{2}{3}x^3\right)-\left[\dfrac{5}{\cancel4_1}x^2·\cancel8^2x-\dfrac{1}{2}x^3\right]-x^3 =$
$\small = \dfrac{5}{4}x^3+\dfrac{2}{3}x^3-\left[5x^2·2x-\dfrac{1}{2}x^3\right]-x^3 =$
$\small = \dfrac{15+8}{12}x^3-\left[10x^3-\dfrac{1}{2}x^3\right]-x^3 =$
$\small = \dfrac{23}{12}x^3-\left[\dfrac{20-1}{2}x^3\right]-x^3 =$
$\small = \dfrac{23}{12}x^3-\dfrac{19}{2}x^3-x^3 =$
$\small = \dfrac{23-114-12}{12}x^3 =$
$\small = -\dfrac{103}{12}x^3 $
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$\small \left[\left(-2abx\right)\left(bx-\dfrac{1}{2}bx\right)+ab^2x^2\right]\left(-a\right)+\left[\left(3a-\dfrac{a}{2}\right)\left(-bx\right)-\dfrac{1}{3}abx\right]\left(-2abx\right) =$
$\small = \left[\left(-2abx\right)\left(\dfrac{2-1}{2}bx\right)+ab^2x^2\right]\left(-a\right)+\left[\left(\dfrac{6-1}{2}a\right)\left(-bx\right)-\dfrac{1}{3}abx\right]\left(-2abx\right) =$
$\small = \left[\left(-\cancel2^1abx\right)\left(\dfrac{1}{\cancel2_1}bx\right)+ab^2x^2\right]\left(-a\right)+\left[\left(\dfrac{5}{2}a\right)\left(-bx\right)-\dfrac{1}{3}abx\right]\left(-2abx\right) =$
$\small = \left[-ab^2x^2+ab^2x^2\right]\left(-a\right)+\left[-\dfrac{5}{2}abx-\dfrac{1}{3}abx\right]\left(-2abx\right) =$
$\small = \left[0\right]\left(-a\right)+\left[\dfrac{-15-2}{6}abx\right]\left(-2abx\right) =$
$\small = 0+\left[-\dfrac{17}{6}abx\right]\left(-2abx\right) =$
$\small = \left[-\dfrac{17}{\cancel6_3}abx\right]\left(-\cancel2^1abx\right) =$
$\small = \dfrac{17}{3}a^2b^2x^2$
-x^3y - 4/3x^3y - 9/4x^3y + 2x^3y + x^3y/4
x^3y(-1 - 4/3 - 9/4 + 2 + 1)
x^3y(-12-16-27+24+3)/12
-28x^3y/12
-7x^3y/3