Question

............... dau coroană cu tot cu rezolvării​

Answer 1

Răspuns:

I. 5

II.24

III. 15,(9)

Explicație pas cu pas:

I. √(25x²+20x+20)=√[5(5x²+4x+4)]=√[5(5x²+4x+4/5+16/5)]=√{5[(√5x+2√5/5)²+16/5]}=√[5(√5x+2√5/5)²+16]≥√16=4 ⇒ √(25x²+20x+20)≥4

√(y²-6√2y+19)=√(y²-6√2y+18+1)=√[(y-3√2)²+1]≥√1=1

⇒ a+b≥4+1

II. √(4x²+8x+85)=√[(4x²+8x+4+81)]=√[(2x+2)²+81]≥√81=9

    √(5y²+10y+69)=√[(5y²+10y+5+64)]=√[(√5y+√5)²+64≥√64=8

     √(6x²+12x+55)=√[(6x²+12x+6+49)=√[(√6x+√6)²+49]≥√49=7

    a+b+c≥9+8+7=24

III. √(x²+4x+20)=√[(x²+4x+4+16)=√[(x+2)²+16]≥√16=4

    √(4y²+8y+29)=√[(4y²+8y+4+25)]=√[(2y+2)²+25]≥√25=5

     √(9z²+12z+53)=√[(9z²+12z+4+49)]=√[(3z+2)²+49]≥√49=7

   a+b+c≥16 ⇒ a+b+c>15,(9)