A ={ 0;-4;25,-64;0,36;0,4;3(27);6}.
Daca A={0, -4, 25; -64, 36/100; 4/10; 324/99; 6
B={x∈Q | x²∈A} =>
B={ 5/1 ; 6/10}
3)Daca triunghiul ABC =echilateral de latura 24 cm si
MA= MB=MC=12√3 cm=> MABC=piramida triunghiulara regulata
a) daca D apartine (BC) astfel incat [BD] congruent cu [DC], calculatia aria triunghiului MAD .b) daca MN perpendicular AD , N apartine ( AD) , calculati lungimea segmentului MN
In ΔMBC , daca BD=DC si MB=MC=12√3, atunci ΔMBC este isoscel=> MD_|_BC
MD²=MB²-BD²=12²*3-12²=12²*2
MD=12√2
in Δechilateral ABC, AD=AB√3/2= 24√3/2=12√3
Ducem MO inaltimea piramidei=> MN_|_AD
ND=1/3*AD=4√3
MN²=MD²-ND²=288-48=240
MN=4√15
Aria ΔMAD=AD*MN/2=12√3*4√15/2=72√5
