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(2x/x+1)+(5/2x)=2
We move all terms to the left:
(2x/x+1)+(5/2x)-(2)=0
Domain of the equation: x+1)!=0
x∈R
Domain of the equation: 2x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
(2x/x+1)+(+5/2x)-2=0
We get rid of parentheses
2x/x+5/2x+1-2=0
We calculate fractions
4x^2/2x^2+5x/2x^2+1-2=0
We add all the numbers together, and all the variables
4x^2/2x^2+5x/2x^2-1=0
We multiply all the terms by the denominator
4x^2+5x-1*2x^2=0
Wy multiply elements
4x^2-2x^2+5x=0
We add all the numbers together, and all the variables
2x^2+5x=0
a = 2; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·2·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*2}=\frac{-10}{4} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*2}=\frac{0}{4} =0 $
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