Question

Va rog foarte frumos, am probleme la algebra. Nu inteleg nimic. Puteti sa-mi explicati cateva dintre exercitiile acestea? ​

Answer 1

Răspuns:

Explicație pas cu pas:

Revin cu restul

Answer 2

$\it I)\\ \\ 1.\ \ 75-3(x-1)^2=0|_{:3} \Rightarrow 25-(x-1)^2=0 \Rightarrow (x-1)^2=25 \Rightarrow \\ \\ \Rightarrow \sqrt{(x-1)^2}=\sqrt{25} \Rightarrow |x-1|=5 \Rightarrow x-1=\pm5 \Rightarrow x-1\in\{-5,\ 5\}|_{+1} \Rightarrow \\ \\ \Rightarrow x\in\{-4,\ 6\}$

Ecuația admite două soluții:

$\it x_1=-4,\ \ x_2=6$

Suma celor două soluții este:

$\it x_1+x_2=-4+6=2$

$\it 2.\ \ (\sqrt{98}+\sqrt{50}-\sqrt{18}):\sqrt2=\sqrt{98}:\sqrt2+\sqrt{50}:\sqrt2-\sqrt{18}:\sqrt2=\\ \\ =\sqrt{98:2}+\sqrt{50:2}-\sqrt{18:2}=\sqrt{49}+\sqrt{25}-\sqrt9=7+5-3=9$

$\it 4.\\ \\ \\ \left.\begin{aligned}\it a= \dfrac{1}{\sqrt2} \Rightarrow a^2=\dfrac{1}{2}<1\\ \\ \\ \it b=\dfrac{2}{\sqrt3}\Rightarrow b^2=\dfrac{4}{3}>0\end{aligned}\right\} \Rightarrow \dfrac{2}{\sqrt3}>\dfrac{1}{\sqrt2}$

$\it 5.\\ \\ \sqrt{(\sqrt3-2)^2}+\sqrt3= |\underbrace{\sqrt3-2}_{<0}|+\sqrt3=-\sqrt3+2+\sqrt3=2$