ko bt lm :<<

\(\dfrac{x}{z}=\dfrac{z}{y}\Rightarrow xy=z^2\)

\(VT=\dfrac{x^2+z^2}{y^2+z^2}=\dfrac{x^2+xy}{y^2+xy}=\dfrac{x\left(x+y\right)}{y\left(x+y\right)}=\dfrac{x}{y}=VP\)

\(\overline{ab}+\overline{ba}=11\left(a+b\right)\)

11 là số nguyên tố để 11(a+b) là số chính phương thì a+b=11

\(\Rightarrow\overline{ab}=\left\{29;38;47;56;65;74;83;92\right\}\)

a: xN,xO,xO,xM,NO,NM,Nx,My,MO

b: ON và OM

c; NO và Nx

d: Mx và My

e: Còn gọi là tia OM

1 A

2 C

C

C.swimming

Đề đâu mà giúp :P

???

\(\sqrt{\left(24+8\sqrt{5}\right)^2}-\sqrt{\left(9-4\sqrt{5}\right)^2}=24+8\sqrt{5}-9+4\sqrt{5}=15+12\sqrt{5}\)

\(\sqrt{\left(17-12\sqrt{2}\right)^2}+\sqrt{\left(9+4\sqrt{2}\right)^2}=17-12\sqrt{2}+9+4\sqrt{2}=26-8\sqrt{2}\)

\(\sqrt{\left(6-4\sqrt{2}\right)^2}+\sqrt{\left(22-12\sqrt{2}\right)^2}=6-4\sqrt{2}+22-12\sqrt{2}=28-16\sqrt{2}\)

\(ô,\\ \Rightarrow24+8\sqrt{5}-\sqrt{\left(9-4\sqrt{5}\right)^2}\\ \Rightarrow24+8\sqrt{5}-\left(9-4-\sqrt{5}\right)\\ \Rightarrow24+8\sqrt{5}-9+4\sqrt{5}\\ \Rightarrow15+8\sqrt{5}+4\sqrt{5}\\ \Rightarrow15+12\sqrt{5}\) 

\(ơ,\\ g\left(17-12\sqrt{2}\right)+\sqrt{\left(9+4\sqrt{2}\right)^2}\\ \Rightarrow g\left(17-12\sqrt{2}\right)+\sqrt{\left(9+4+\sqrt{2}\right)^2}\\ \Rightarrow\left(17-12\sqrt{2}\right)g+\sqrt{\left(9+4\sqrt{2}\right)^2}\\ \Rightarrow\left(17-12\sqrt{2}\right)g+9+4\sqrt{2}\) 

\(u,\\ 6-4\sqrt{2}+\sqrt{\left(22-12\sqrt{2}\right)}^2\\ \Rightarrow6-4\sqrt{2}+22-12\sqrt{2}\\ \Rightarrow28-4\sqrt{2}-12\sqrt{2}\\ \Rightarrow28-16\sqrt{2}\)

ô \(\sqrt{\left(24+8\sqrt{5}\right)^2}-\sqrt{\left(9-4\sqrt{5}\right)^2}\)

= 24 + 8\(\sqrt{5}\) - 9 + 4\(\sqrt{5}\)

= 15 + 12\(\sqrt{5}\)

ơ \(\sqrt{\left(17-12\sqrt{2}\right)^2}+\sqrt{\left(9+4\sqrt{2}\right)^2}\)

= 17 - 12\(\sqrt{2}\) + 9 + 4\(\sqrt{2}\)

= 26 - 8\(\sqrt{2}\)

u \(\sqrt{\left(6-4\sqrt{2}\right)^2}+\sqrt{\left(22-12\sqrt{2}\right)^2}\)

= 6 - 4\(\sqrt{2}\)  + 22 - 12\(\sqrt{2}\)

= 28 -16\(\sqrt{2}\)