Question

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Answer 1

c) inmultim de diagonala
Conditii de existenta
3x-2 != 0 => x != 3/2
3x+1 != 0 => x != -1/3 (!= inseamna diferit)
(2x+1)(3x+1)=(3x-2)(2x+5)
6x²+2x+3x+1=6x²+15x-4x-10
5x+1=11x-10
-6x=-11
x=11/6
d) C.e. 2x-3 !=0 => x != 3/2
2(2x-3)=(2√3-√2)(√2+√12)
4x-6 = 2√6+2√36-2-√24
4x-6 = 2√6+12-2-2√6
4x-6 = 10
4x = 16
x = 4

Answer 2

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