URAIKAN BENTUK ( 2x - 3Y )^5 dan ( x + 1per x ) ^5 ?.... Question from @nagitawasya

October 2018 1 67 Report

URAIKAN BENTUK ( 2x - 3Y )^5 dan ( x + 1per x ) ^5 ?

Takamori37 Segitiga pascal untuk pangkat 5 adalah:
1,5,10,10,5,1
Sehingga,

a.)
$$\begin{align}&(2x-3y)^5 \\&=(2x)^5+5(2x)^4(-3y)^1+10(2x)^3(-3y)^2+10(2x)^2(-3y)^3+\ddots \\ &\qquad 5(2x)^1(-3y)^4+(-3y)^5 \\ &=32x^5-15y.16x^4+10.8x^3.9y^2+10(4x^2)(-27y^3)+10x(81y^4)-243y^5 \\ &=32x^5-240x^4y+720x^3y^2-810x^2y^3+810xy^4-243y^5\end{align}$

b.)
$$\begin{align}&(x+\frac{1}{x})^5 \\ &=(x+x^{-1})^5 \\ &=x^5+5x^4(x^{-1})+10x^3(x^{-2})+10x^2(x^{-3})+5x(x^{-4})+x^{-5} \\ &=x^5+5x^3+10x+10x^{-1}+5x^{-3}+x^{-5} \\ &=x^5+5x^3+10x+\frac{10}{x}+\frac{5}{x^3}+\frac{1}{x^5}\end{align}$

0 votes Thanks 1

nagitawasya tolongin lagi yah 3 soal lagi kok

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