Question

Va rog mult de tot este urgent!!!!!! 1. Sa se calculeze
2√2*[2√243+2*(4√32-3√27)]
2.Aflati x in proportile:
a)×/2√3=2√3-3/√3
b)√13+√1296 radicalul din fata este si peste √1296/x=3^3-3^2-3-3^0/√23+√1681 √ de la 23 este si peste√1681

Answer 1

2√2*[2√243+2*(4√32-3√27)]=2√2*[2*9√3+2*(4*4√2-3*3√3)]=2√2*[18√3+2*(16√2-9√3)]=2√2*(18√3+32√2-18√3)=2√2*32√2=2*32*2=128


×/2√3=2√3-3/√3
x=2√3*2√3-3*2
x=2²·3-6
x=6

√13+√1296/x=3³-3²-3-3°/√23+√1681
√13+36/x=27-9-3-1/√23+41
√49/x=14/√64
7/x=14/8
x=7·8/14
x=56/14
x=4