Question

Va rog sa ma ajutati!!

1. Determinati x ∈ R pentru care 4(2x-5)²-(3x+1)²=0

2. Rezolvati ecuatia
$3(2x+1)^{2} = \sqrt{( \sqrt{5-3)^{2} } } + \sqrt{21+4 \sqrt{5} } - \sqrt{( \sqrt{5-2)^{2} } }$

Answer 1

1.  4(4x² - 20x + 25) - 9x² - 6x - 1 = 0
16x² - 80x + 100 - 9x² - 6x -1 = 0
7x² - 86x + 99 = 0
7x² - 77x - 9x +99 = 0    7x(x - 11) - 9(x-11) = 0    (x-11)(7x - 9) = 0  x1 = 11  x2 =9/7
2. 3(4x² + 4x + 1) = √2 + √20 +1 - √3
√(a+√b) = √[a+√(a² -b)]/2 + √[a - √(a² - b)]/2
√(21+4√5) = √[21 + √(441 - 80)]/2 - √[21 - √(441 - 80)]/2 =
= √(21+ 19)/2 + √(21- 19)/2 = √20 + 1
12x² +12x + 3 = √2 + 2√5 - √3 + 1
12x(x+1) = √2 +2√5 - √3 - 2
x(x+1) = (√2 + 2√5 - √3 - 2)/12....... cred ca  ceva e scris gresit.....