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[sqrt(9x^(2)+6x+1)<2-x" then "],[x in(-(...

[sqrt(9x^(2)+6x+1)<2-x" then "],[x in(-(3)/(2),(1)/(4))quad " 2) "x in(-(3)/(2),(1)/(4)]],[x in[-(3)/(2),(1)/(4))quad " 4) "x<(1)/(4)]

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Using properties of proportion, solve for x : (i) (sqrt(x + 5) + sqrt(x - 16))/ (sqrt(x + 5) - sqrt(x - 16)) = (7)/(3) (ii) (sqrt(x + 1) + sqrt(x - 1))/ (sqrt(x + 1) - sqrt(x - 1)) = (4x -1)/(2) . (iii) (3x + sqrt(9x^(2) -5))/(3x - sqrt(9x^(2) -5)) = 5 .

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