I need help how to solve it mot the answer plz and thank u

Mathematics

2 answers:

jek_recluse [69]3 years ago

4 0

The second one would be the answer. You want to plug in for this question. You can divide over the five leaving you with (x-2)^2 =25. Since you are squaring this answer, you want to get either 5 or negative five. Both -3 and 7 will do that.

Tatiana [17]3 years ago

4 0

(x-2) ^2 = 125/5,
from formula (a-b)^2= a^2 - 2ab+ b^2
the equation becomes
x^2 - 4x+4= 25
x^2-4x-21= 0
x^2 -7x+3x-21=0
x(x-7)+3(x-7)= 0
(x+3) (x-7)= 0
so either x+3=0
or x-7=0
x can have values x= -3, 7

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Shkiper50 [21]

Answer:

$x=1$

Step-by-step explanation:

$~~~~~x = \dfrac{1}{2 - \dfrac 1{2-\dfrac{1}{2-x}}}\\\\\\\implies x = \dfrac{1}{2 - \dfrac 1{\dfrac{4-2x-1}{2-x}}}\\\\\\\implies x= \dfrac{1}{2 - \dfrac{2-x}{3-2x}}\\\\\\\implies x = \dfrac{1}{\dfrac{6-4x-2+x}{3-2x}}\\\\\\\implies x = \dfrac{3-2x}{4-3x}\\\\\\\implies 4x -3x^2 = 3-2x\\\\\\\implies 3x^2 -4x +3-2x = 0\\\\\\\implies 3x^2 -6x +3 = 0\\\\\\\implies 3(x^2 -2x +1) =0\\\\\\\implies x^2 -2x +1 = 0\\\\\\\implies (x-1)^2 = 0\\\\\\\implies x -1 = 0\\\\\\\implies x = 1$