Question

Допомжіть, помогитееееее, пжжж​

Answer 1

Ответ:

$1)\ \ \dfrac{2x^2-1}{x^2-9}-\dfrac{x+1}{x+3}=\dfrac{3x+1}{3x-9}\ \ ,\ \ \ ODZ:\ x\ne \pm 3\\\\\\\dfrac{2x^2-1}{(x-3)(x+3)}-\dfrac{x+1}{x+3}-\dfrac{3x+1}{3(x-3)}=0\\\\\\\dfrac{3(2x^2-1)-3(x-3)(x+1)-(x+3)(3x+1)}{3(x-1)(x+1)}=0\\\\\\\dfrac{6x^2-3-3(x^2-2x-3)-(3x^2+10x+3)}{3(x-3)(x+3)}=0\\\\\\\dfrac{-4x+3}{3(x-3)(x+3)}=0\ \ ,\ \ -4x+3=0\ \ ,\ \ 4x=3\ \ ,\ \ \underline {\ x=0,75\ }$

$2)\ \ -2\cdot \Big(x^2+\dfrac{1}{x^2}\Big)+9\cdot \Big(x+\dfrac{1}{x}\Big)-14=0\ \ ,\ \ \ ODZ:\ x\ne 0\\\\\\(x+\dfrac{1}{x})=t\ \,\ \ t^2=x^2+\dfrac{1}{x^2}+2\ \ \to \ \ (x^2+\dfrac{1}{x^2})=t^2-2\\\\\\-2\cdot (t^2-2)+9t-14=0\ \ ,\ \ -2t^2+4+9t-14=0\ \ ,\\\\2t^2-9t+10=0\ \ ,\ \ D=1\ \ ,\ \ t_1=\dfrac{9-1}{4}=2\ \ ,\ \ t_2=\dfrac{5}{2}\\\\\\a)\ \ x^2+\dfrac{1}{x^2}=2\ ,\ \ \dfrac{x^4-2x^2+1}{x^2}=0\ ,\ \ x^4-2x^2+1=0\ ,\ (x^2-1)^2=0\ ,\\\\\\x^2-1=0\ \ ,\ \ x^2=1\ \ ,\ \ \underline {\ x=\pm 1\ }$

$b)\ \ x^2+\dfrac{1}{x^2}=\dfrac{5}{2}\ \ ,\ \ \dfrac{2x^4-5x^2+2}{2x^2}=0\ ,\ 2x^4-5x^2+2=0\ ,\ \ D=9\ ,\\\\\\\underline {\ x_1=\dfrac{5-3}{4}=\dfrac{1}{2}=0,5\ \ ,\ \ \ x_2=2\ }\\\\\\Otvet:\ \ x=-1\ ,\ x=0,5\ ,\ x=1\ ,\ x=2\ .$

$3)\ \ \sqrt[3]{x+7}-\sqrt{x+3}=0\ \ ,\ \ \ \ ODZ:\ x\geq -3\ ,\\\\\sqrt[3]{x+7}=\sqrt{x+3}\ \ \to \ \ \ \sqrt[6]{(x+7)^2}=\sqrt[6]{(x+3)^3}\ \Big|\ \uparrow 6\\\\(x+7)^2=(x+3)^3\\\\x^2+14x+49=x^3+9x^2+27x+27\\\\x^3+8x^2+13x-22=0\\\\x=1\ \ \to \ \ 1^3+8\cdot 1^2+13\cdot 1-22=1+8+13-22=22-22=0\ \ \to \\\\x^3+8x^2+13x-22=(x-1)(x^2+9x+22)\\\\(x-1)(x^2+9x+22)=0\\\\a)\ \ x-1=0\ ,\ \ \underline {\ x=1\ }\\\\b)\ \ x^2+9x+22=0\ \ ,\ \ D=-7<0\ \ \to \ \ x\in \varnothing \\\\Otvet:\ \ x=1\ .$