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Question Number 103685 by bemath last updated on 16/Jul/20

	Answered by bobhans last updated on 16/Jul/20
	$x−(√((2−x)(3−x))) = (√(4−x)){(√(3−x))+(√(2−x))} 2x^2 −5x+6−2x(√((2−x)(3−x))) = (4−x)(5−2x)+2(√((3−x)(2−x))) 4x−7 = x(√((2−x)(3−x)))+(4−x)(√((2−x)(3−x))) 4x−7 = 4(√((2−x)(3−x))) (4x−7)^2 = 16(6−5x+x^2 ) −56x+49 = 96−80x 24x = 47 ⇒ x = ((47)/(24)) ★$
	$x - \sqrt{(2 - x) (3 - x)} = \sqrt{4 - x} {\sqrt{3 - x} + \sqrt{2 - x}}$ $2 x^{2} - 5 x + 6 - 2 x \sqrt{(2 - x) (3 - x)} =$ $(4 - x) (5 - 2 x) + 2 \sqrt{(3 - x) (2 - x)}$ $4 x - 7 =$ $x \sqrt{(2 - x) (3 - x)} + (4 - x) \sqrt{(2 - x) (3 - x)}$ $4 x - 7 = 4 \sqrt{(2 - x) (3 - x)}$ ${(4 x - 7)}^{2} = 16 (6 - 5 x + x^{2})$ $- 56 x + 49 = 96 - 80 x$ $24 x = 47 \Rightarrow x = \frac{47}{24} ★$
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