Question

Срочно пожалуйста даю 60 баллов ​

Answer 1

Ответ:

$1)\ \ \Big(\dfrac{6}{5}+1\dfrac{1}{10}\Big)\cdot 24=\dfrac{12+11}{10}\cdot 24=\dfrac{23\cdot 12}{5}=55,2\\\\\\2)\ \ 3\sqrt{14}\approx11,225\ \ ,\ \ 5\sqrt5\approx 11,180\ \ ,\ \ 2\sqrt{31}\approx 11,136\\\\2\sqrt{31}<5\sqrt5<3\sqrt{14}\\\\\\3)\ \ 2x^2+x-5=0\ \ ,\ \ D=41\ \ ,\ \ x_{1}=\dfrac{-1-\sqrt{41}}{4}\ \ ,\ \ \ x_2=\dfrac{-1+\sqrt{41}}{4}$

$4)\ \ 3\sqrt2\cdot \sqrt5\cdot 4\sqrt{10}=12\cdot \sqrt{2\cdot 5\cdot 10}=12\cdot \sqrt{100}=12\cdot 10=120$

$5)\ \ 3x^2+bx+24=0\ \ ,\ \ x_1=8\\\\teorema\ Vieta:\ \left\{\begin{array}{l}x_1\cdot x_2=24\\x_1+x_2=-b\end{array}\right\ \ \left\{\begin{array}{l}8\cdot x_2=24\\8+x_2=-b\end{array}\right\ \ \left\{\begin{array}{l}x_2=3\\8+3=-b\end{array}\right\\\\\\\left\{\begin{array}{l}x_2=3\\b=-11\end{array}\right$

$6)\ \ \dfrac{4x+1}{x-3}=\dfrac{3x-8}{x+1}\ \ ,\ \ \ \ ODZ:\ x\ne 3\ ,\ x\ne -1\ ,\\\\\\\dfrac{(4x+1)(x+1)-(3x-8)(x-3)}{(x-3)(x+1)}=0\\\\\\4x^2+5x+1-(3x^2-17x+24)=0\\\\x^2+22x-23=0\ \ ,\ \ x_1=-23\ ,\ x_2=1\ \ (teorema\ Vieta)$