(7-2x)(49+14x4x^2)+2x(2x-5)(2x+5)=43 100(0,2x+1)(0,04x^2-0,2x+1)=5x(0,16x^2-4). Created by DimaTsarelynga. algebra-ru

(7-2x)(49+14x4x^2)+2x(2x-5)(2x+5)=43 100(0,2x+1)(0,04x^2-0,2x+1)=5x(0,16x^2-4)...

July 2022 2 4 Report

(7-2x)(49+14x4x^2)+2x(2x-5)(2x+5)=43
100(0,2x+1)(0,04x^2-0,2x+1)=5x(0,16x^2-4)

kik51
$100(0.2x + 1)(0.04x {}^{2} - 0.2x + 1) = 5x(0.16x {}^{2} - 4) \\ \frac{4x {}^{3} }{5} + 100 = \frac{4}{5} x {}^{3} - 20x \\ 4x {}^{3} + 500 = 4x {}^{3} - 100x \\ 4x {}^{3} + 500 - 4x {}^{3} + 100x = 0 \\ 100x + 500 = 0 \\ 100x = - 500 \\ x = - 5$

3 votes Thanks 1

sedinalana (7-2x)(49+14x+4x^2)+2x(2x-5)(2x+5)=43
343-8x³+8x³-50x=43
50x=343-43
50x=300
x=300:50
x=6

100(0,2x+1)(0,04x^2-0,2x+1)=5x(0,16x^2-4)
100(0,008x³+1)=5x(0,16x²-4)
0,8x³+100-0,8x³+20x=0
20x=-100
x=-100:20
x=-5

6 votes Thanks 5

Answers & Comments

element nahoditsya v 4 gruppe periodicheskoj sistemy massovaya dolya elementa v eg

23 zadanie pozhalujsta6b707eba191f9a64fea6230354efd321 60929

u yakomu roci shvejcariya zdobula avtonomiyu u skladi syashenoj rimskoj imperii

рекомендуемые вопросы

Helpful Links

Helpful Social