Question

Pls va roggggg.....!!!!​

Answer 1

Răspuns:

Sper ca înțelegi din poză

Answer 2

$\bf {5.} \it \ 3+2\sqrt2=2=1+2\sqrt2=(\sqrt2)^2+2\sqrt2+1^2=(\sqrt2+1)^2\\ \\ a=\sqrt{(\sqrt2+1)^2}=\sqrt2+1\\ \\ Analog,\ \ b=\sqrt2-1\\ \\ a-b=\sqrt2+1-(\sqrt2-1)=\sqrt2+1-\sqrt2+1=2\Rightarrow (a-b)^2=2^2=4$

3.

Fie x < y < z măsurile celor trei unghiuri.

$\it Fie\ x< y<z\ -\ m\breve asurile\ celor\ trei\ unghiuri.\\ \\ x+y+z=180^o \Rightarrow x=180^o-(y+z)\ \ \ \ \ \ \ (*)\\ \\ \{x,\ y,\ z\}\ d.\ p.\ \{4,\ 5,\ 6\}\Rightarrow \dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y+z}{4+5+6}=\dfrac{180^o}{15}=12^o\\ \\ \\ \dfrac{x}{4}=12^o\Rightarrow x=4\cdot12^o\Rightarrow x=48^o\ \stackrel{(*)}{\Longrightarrow}\ 180^o-(y+z)=48^o\Rightarrow \\ \\ \\ \Rightarrow y+z=180^o-48^o=132^o$