Question

Срочно!!! Помогите решить пожалуйста.
1) 9^x-23*3^((x-3)/2)=4*3^(-x)
2) 25^sin^2x+100=5*(1/5)^cos^2x-2

Answer 1

$1)quad 9^{x}-23cdot 3^{frac{x-3}{2}}=4cdot 3^{-x}\\3^{2x}-23cdot 3^{frac{x}{2}}cdot 3^{-frac{3}{2}}-4cdot frac{1}{3^{x}}=0\\t=3^{frac{x}{2}} textgreater 0; ,; ; t^4-23cdot tcdot frac{1}{sqrt{3^3}}-frac{4}{t^2}=0; |cdot sqrt{3^3}t^2\\sqrt{3^3}t^6-23t^3-4sqrt{3^3}=0\\z=t^3 textgreater 0; ,; 3sqrt3z^2-23z-12sqrt3=0\\D=23^2-4cdot 3sqrt3(-12sqrt3)=961; ; ,; ; sqrt{961}=31$

$z_1=frac{23-31}{6sqrt3}=-frac{8}{6sqrt3}=-frac{4}{2sqrt3} textless 0; ,; ; z_2=frac{54}{6sqrt3}}=frac{9}{sqrt3}=3^{3/2}$

$z=t^3=3^{frac{3x}{2}}=3^{frac{3}{2}}; ; Rightarrow ; ; ; frac{3x}{2}=frac{3}{2}\\underline{x=1}$

$2)quad 25^{sin^2x}+100=5cdot (frac{1}{5})^{cos^2x-2}\\5^{2sin^2x}+(5cdot 2)^2-5cdot 5^{2-cos^2x}=0\\5^{2sin^2x}+5^2cdot 2^2-5^{2+(1-cos^2x)}=0; ,; ; ; (1-cos^2x=sin^2x)\\(5^{sin^2x})^2-5^2cdot 5^{sin^2x}+5^2cdot 2^2=0\\t=5^{sin^2x} textgreater 0; ,; ; ; t^2-25t+100=0\\D=25^2-4cdot 100=225; ,; ; sqrt{225}=15\\t_1=frac{25-15}{2}=5 textgreater 0; ,; ; t_2=20 textgreater 0\\a); ; 5^{sin^2x}=5; ; to ; ; sin^2x=1; ,; ; frac{1-cos2x}{2}=1; ,; cos2x=-1; ,$

$2x=pi +2pi n,; nin Z\\x=frac{pi}{2}+pi n,; nin Z$

$b); ; 5^{sin^2x}=20\\sin^2x=log_520=log_5(5cdot 2^2)=1+2log_52 textgreater 1\\Tak; kak; ; 0 leq sin^2x leq 1; ,; to; ; sin^2x=1+2log_52; ; ne; ; imeet; ; reshenij\\Otvet:; ; x=frac{pi }{2}+pi n,; nin Z$