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mi servono i numeri 119 120 e 121 grazie mille in anticipo

IMG 2935

 

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@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.

@fagottino un esercizio per volta, foto leggibile. Ciao buona domenica

5 Risposte



3
119

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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 $

@gramor 👍👌👍

@remanzini_rinaldo - Grazie mille, saluti.



3
121

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$\small \left[\left(-2abx\right)\left(bx-\dfrac{1}{2}bx\right)+ab^2x^2\right]\left(-a\right)+\left[\left(3a-\dfrac{a}{2}\right)\left(-bx\right)-\dfrac{1}{3}abx\right]\left(-2abx\right) =$

$\small = \left[\left(-2abx\right)\left(\dfrac{2-1}{2}bx\right)+ab^2x^2\right]\left(-a\right)+\left[\left(\dfrac{6-1}{2}a\right)\left(-bx\right)-\dfrac{1}{3}abx\right]\left(-2abx\right) =$

$\small = \left[\left(-\cancel2^1abx\right)\left(\dfrac{1}{\cancel2_1}bx\right)+ab^2x^2\right]\left(-a\right)+\left[\left(\dfrac{5}{2}a\right)\left(-bx\right)-\dfrac{1}{3}abx\right]\left(-2abx\right) =$

$\small = \left[-ab^2x^2+ab^2x^2\right]\left(-a\right)+\left[-\dfrac{5}{2}abx-\dfrac{1}{3}abx\right]\left(-2abx\right) =$

$\small = \left[0\right]\left(-a\right)+\left[\dfrac{-15-2}{6}abx\right]\left(-2abx\right) =$

$\small = 0+\left[-\dfrac{17}{6}abx\right]\left(-2abx\right) =$

$\small = \left[-\dfrac{17}{\cancel6_3}abx\right]\left(-\cancel2^1abx\right) =$

$\small = \dfrac{17}{3}a^2b^2x^2$

@gramor 👍👌👍

@remanzini_rinaldo - Grazie mille, saluti.



2
image

xy^2/4+3xy^2/4+3/2xy^2-1/2xy^2

xy^2+xy^2

2xy^2

 

 



2
image

-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



2
image

-2ab(-4a^2b^2 + 3a^2b^2)

-2ab*-a^2b^2

2a^3b^3



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