Toán Lớp 9: a) √-3x+7 b)1/ √9-12x+4x^2 c) √x+2/4-x d) √x+1/5x+2

Question

Toán Lớp 9:  a) √-3x+7 b)1/ √9-12x+4x^2 c) √x+2/4-x d) √x+1/5x+2

tuyen98 · Answer

~rai~ (a) sqrt{-3x+7} ĐKXĐ:-3x+7 ge 0 Leftrightarrow -3x ge -7 Leftrightarrow x< dfrac{7}{3}. b) dfrac{1}{ sqrt{9-12+4x^2}} ĐKXĐ:9-12x+4x^2>0. quad (1) text{Ta có:}9-12x+4x^2=(3-2x)^2 ge 0 quad forall x in mathbb{R} quad nên (1) Leftrightarrow 9-12x+4x^2 ne 0 (3-2x)^2 ne 0 Leftrightarrow 3-2x ne 0 Leftrightarrow 2x ne 3 Leftrightarrow x ne dfrac{3}{2}. c) sqrt{ dfrac{x+2}{4-x}} ĐKXĐ: dfrac{x+2}{4-x} ge 0 Leftrightarrow left[ begin{array}{I} begin{cases}x+2 ge 0 4-x>0 end{cases} begin{cases}x+2 le 0 4-x<0 end{cases} end{array} right. Leftrightarrow left[ begin{array}{I} begin{cases}x ge -2 x<4 end{cases} begin{cases}x le -2 x>4 end{cases} text{(vô lí)} end{array} right. Leftrightarrow -2 le x<4. d) sqrt{ dfrac{x+1}{5x+2}} ĐKXĐ: dfrac{x+1}{5x+2} ge 0 Leftrightarrow left[ begin{array}{I} begin{cases}x+1 ge 0 5x+2>0 end{cases} begin{cases}x+1 le 1 5x+2<0 end{cases} end{array} right. Leftrightarrow left[ begin{array}{I} begin{cases}x ge -1 x>- dfrac{2}{5} end{cases} begin{cases}x le -1 x<- dfrac{2}{5} end{cases} end{array} right. Leftrightarrow left[ begin{array}{I}x>- dfrac{2}{5} x le -1. end{array} right. )