Va rog doar 4 am nevoie urgeenttt.Dau coroanaaaaa

a)Ma(x,y)=[tex] \frac{x+y}{2}= \frac{2.4+15}{2}= \frac{17.4}{2}=8.7 [/tex]
Mg(x,y)=[tex] \sqrt{x*y}= \sqrt{2.4*15}= \sqrt{36}=6 [/tex]
b)Ma(x,y)=[tex] \frac{ 2\sqrt{6}+3 \sqrt{6} }{2}= \frac{ 5\sqrt{6} }{2} [/tex]
Mg(x,y)=[tex] \sqrt{x*y}= \sqrt{ 2\sqrt{6} *3 \sqrt{6} }=6 \sqrt{6} [/tex]
c)
x=I3-2√3I=2√3-3
3 < 2√3
9 < 4*3
9 < 12
Ma(x,y)=[tex] \frac{x+y}{2}= \frac{ 2\sqrt{3}-3+3+2 \sqrt{3} }{2}= \frac{ 4\sqrt{3} }{2}= 2\sqrt{3} [/tex]
Mg(x,y)=[tex] \sqrt{( 3^{2}-12 }= \sqrt{9-12}= \sqrt{-3} [/tex]

4) a) x=2,4 sau 12/5 si y=15
ma=(x+y)/2
ma=(2,4+15)/2
ma=17,2/2
ma=8,7
mg=√(a•b)
mg=√(12/5•15)
mg=√12•3
mg=√36
mg=6
b) x=2√6 si y=3√6
ma=(a+b)/2
ma=(2√6+3√6)/2
ma=5√6/2
mg=√a•b
mg=√(2√6•3√6)
mg=√6•6
mg=√36
mg=6
c) x= |3–2√3| si y=3+2√3
x=2√3–3
ma=(x+y)/2
ma=(2√3–3+3+2√3)/2
ma=4√3/2
ma=2√3
mg=√a•b
mg=√(2√3–3)(3+2√3)
mg=√(6√3+12–9–6√3)
mg=√3
d) x=√2+1+1/√2–1 si y=√2–1+1/√2+1
x=√2+1+√2+1
x=2√2+2
y=√2–1+√2–1
y=2√2–2
ma=(x+y)/2
ma=(2√2+2+2√2–2)/2
ma=4√2/2
ma=2√2
mg=√a•b
mg=√(2√2+2)(2√2–2)
mg=√8–4
mg=√4
mg=2