Question

Срочно заранее спасибо

Answer 1

$1); ; 16- sqrt{frac{2}{3}x} =12; ,; ; ; ODZ:; ; x geq 0\\sqrt{frac{2}{3}x}=4; ,; ; frac{2}{3}x=16; ,; ; underline {x=24}\\2); ; 1+x=sqrt{1+xcdot sqrt{x^2+12}}; ,; ; ; ODZ:; ; 1+x geq 0; ,; ; x geq -1\\1+2x+x^2=1+xcdot sqrt{x^2+12}\\xcdot sqrt{x^2+12}=2x+x^2\\x^2(x^2+12)=4x^2+4x^3+x^4\\4x^3-8x^2=0\\4x^2(x-2)=0\\underline {x_1=0; ,; ; x_2=2}$

$3); ; sqrt{12+x}=sqrt{7x+8}-2; ,; ; ; ODZ:; left { {{12+x geq 0} atop {7x+8 geq 0}} right. ; left { {{x geq -12} atop {x geq -frac{8}{7}}} right. ; to ; x geq -frac{8}{7}\\sqrt{12+x} +2=sqrt{7x+8} \\12+x+4sqrt{12+x}+4=7x+8\\4sqrt{12+x}=6x-8; |:2\\2sqrt{12+x}=3x-4\\4cdot (12+x)=9x^2-24x+16\\9x^2-28x-32=0\\D/4=196+9cdot 32=484\\underline {x_1 frac{14-22}{9} =-frac{8}{9} ; ,; ; x_2=frac{14+22}{9} =4}$

$4); ; sqrt{9-x} leq 3; ,$

$left { {{9-x geq 0} atop {9-x leq 3^2}} right. ; ; left { {{-x geq -9} atop {-x leq 0}} right. ; ; left { {{x leq 9} atop {x geq 0}} right. ; ; ; to ; ; ; 0 leq x leq 9\\underline {xin [, 0,9, ]}$