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h)√2x+7≥23-3√2x
i)√2x≥√72-√18


Răspuns :

h)√2x+3√2x>\ 23-7
4√2x>\ 16
x>\ 16/4√2
x>\ 4/√2 (rationalizam cu √2)
x>\ 4√2/2
x>\ 2√2
x € [2√2, + infinit)

i) √2x>\ 6√2-3√2
√2x>\ 3√2
x>\ 3√2/√2
x>\ 3
x € [3, +infinit)