Proszę pomużcie na JUtro!!!!!!!!!!!!!!^-potęga a/b loicznik i mianownik(2-3x)^2(5x-4y)^2(4+8)^2(1/2+.... Question from @herakles380

herakles380 @herakles380

November 2018 1 11 Report

Proszę pomużcie na JUtro!!!!!!!!!!!!!!

^-potęga a/b loicznik i mianownik

(2-3x)^2

(5x-4y)^2

(4+8)^2

(1/2+1/3x)^2

(4-1/4)^2

(2,5+1,5Y)^2

(√2+√3)^2

(√5-√7)^2

(√7+2√2)^2

(√2-2√3)^2

(√5+4√2)^2

(1/√2+1/√8)^2

(3 1/3+2X)^2

(Pi-2)^2

(x+4x)^3

animaldk $Wzory:\\\\(a\pm b)^2=a^2\pm2ab+b^2\\\\a)\ (2-3x)^2=2^2-2\cdot2\cdot3x+(3x)^2=4-12x+9x^2\\\\(5x-4y)^2=25x^2-40xy+14y^2\\\\(4+8)^2=12^2=144\\\\(\frac{1}{2}+\frac{1}{3}x)^2=\frac{1}{4}+\frac{1}{3}x+\frac{1}{9}x^2\\\\(4-\frac{1}{4})^2=(\frac{3}{4})^2=\frac{9}{16}\\\\(2,5+1,5y)^2=6,25+7,5y+2,25y^2$

$(\sqrt2+\sqrt3)^2=2+2\sqrt6+3=5+2\sqrt6\\\\(\sqrt5-\sqrt7)^2=5-2\sqrt{35}+7=12-2\sqrt{35}\\\\(\sqrt7+2\sqrt2)^2=7+4\sqrt{14}+8=15+4\sqrt{14}\\\\(\sqrt2-2\sqrt3)^2=2-4\sqrt6+12=14-4\sqrt6\\\\(\sqrt5+4\sqrt2)^2=5+8\sqrt{10}+32=37+8\sqrt{10}\\\\(\frac{1}{\sqrt2}+\frac{1}{\sqrt8})^2=\frac{1}{2}+\frac{1}{2}+\frac{1}{8}=1\frac{1}{8}$

$(3\frac{1}{3}+2x)^2=(\frac{10}{3}+2x)^2=\frac{100}{9}+\frac{400}{3}x+4x^2\\\\(\pi-2)^2=\pi^2-2\\\\
(x+4x)^3=(5x)^3=125x^3\\\\\\\©DRK$

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Błagam do 7:00 obliczanie!hociaż pare pzykładów blagam!^3 -to do potęgi 3 ,1/2 -to jedna druga ^2 do poęgi2(x+4)^3+(x+4)^2(x+1/3)^3-(x-1/3)^2(2x+3)^3+(2x-3)^3(x^2+x)^3-(x^2-x)^3(5+x^2)(√5-x)(√5+x)(x^2+x+4)(x-2)(x^3+8)(x^2-x+1)(x+1)(1-x^3(2x-1)(1+2x+4x^2+8x^3+16x^-4(x^2-1)(1-x^2+x^4+x^8+x^10+x^12)

Answer

herakles380 November 2018 | 0 Replies

Proszę na jutro Matematyka błagam! dam naj^potęga (2x-2)^35+3x)^3(1-y)^3(3+3x)^3(1-√2)^3(2-√3)^3(4-√5)^3(√2-2)^3(√7+√5)^32√2-1)^3(1+3√3)^3

Answer

Proszę na jutro Matematyka błagam! dam naj^potęga (2x-2)^35+3x)^3(1-y)^3(3+3x)^3(1-√2)^3(2-√3)^3(4-√5)^3(√2-2)^3(√7+√5)^32√2-1)^3(1+3√3)^3

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