Question

Aflati x din egalitate
{3^60*3^40: (3^3)^33*2-5:[8079-2*(2x+2)]}-2019^0=0^1

Answer 1

$\frac{3^6\cdot \:0\cdot \:3^4\cdot \:0}{\left(3^3\right)^3\cdot \:3\cdot \:2}-\frac{5}{\left(8079-2\left(2x+2\right)\right)}-2019^0=0^1\quad :\quad x=2020$

$\frac{3^6\cdot \:0\cdot \:3^4\cdot \:0}{\left(3^3\right)^3\cdot \:3\cdot \:2}-\frac{5}{\left(8079-2\left(2x+2\right)\right)}-2019^0=0^1$

$\frac{3^6\cdot \:0\cdot \:3^4\cdot \:0}{\left(3^3\right)^3\cdot \:3\cdot \:2}-\frac{5}{8079-2\left(2x+2\right)}-2019^0+2019^0=0^1+2019^0$

$\frac{3^6\cdot \:0\cdot \:3^4\cdot \:0}{\left(3^3\right)^3\cdot \:3\cdot \:2}-\frac{5}{8079-2\left(2x+2\right)}=1$$\frac{3^6\cdot \:0\cdot \:3^4\cdot \:0}{\left(3^3\right)^3\cdot \:3\cdot \:2}\cdot \:118098\left(-4x+8075\right)-\frac{5}{-4x+8075}\cdot \:118098\left(-4x+8075\right)=1\cdot \:118098\left(-4x+8075\right)$$-590490=118098\left(-4x+8075\right)$

$18098\left(-4x+8075\right)=-590490$

$\frac{118098\left(-4x+8075\right)}{118098}=\frac{-590490}{118098}$

$-4x+8075=-5$

$-4x+8075-8075=-5-8075$

$-4x=-8080$

$\frac{-4x}{-4}=\frac{-8080}{-4}$

x=2020

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