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2√5 3V BV12X" 180, 2. Calculaţi: a) 2√3-5√3+4√3-7√3; b) √√2-3√5 +3√2-7√5-6√2; c) 1-√2+2√2-3+7-5√2+4√2; d) 2√3+315-3√3)+7√/3 5(3√5-2√/2)-$($√5-3√2)+10√5-4√/2; e) f) (5√3+3√2-3)+(1-√√3+2√2)-(5√2+4√3-2).​

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