Question

VA ROG!!!! AJUTOR!!!
DAU COROANA!!!!

1) Calculati si reduceti termeni asemenea:

a) (2x-5)^2 - (2x-3) (2x+3) +21x=

b) (a+6) (a-6) + (a-6)^2 +14a=

c) (2√3)^2=

d) 3√2 - √8 =

e) √12 +√3 -8√3=

f) √300 + √3 -√27 =

g) √150: √6 =

Answer 1

a)

(2x+5)²-(2x-3)(2x+3)+21x=

x²•4+20x+25-x²•4+9+21x=

x²•4+20x+34-x²•4+21=41x+34

b)

(a+6)(a-6)+(a-6)²+14a=

2a+2a²=2a(1+a)

$c) \\ \\ (2 \sqrt{3}) ^{2} = 4 \times 3 = 12 \\ \\ d) \\ \\ 3 \sqrt{2} - \sqrt{8} = 3 \sqrt{2} - \sqrt{ {2}^{2} \times 2 } \\ \\ 3\sqrt{2} - 2 \sqrt{2} = \sqrt{2} \\ \\ e) \\ \\ \sqrt{12} + \sqrt{3} - 8\sqrt{3} = \sqrt{ {2}^{2} \times 3 } + \sqrt{3} - 8 \sqrt{3} \\ \\ 2 \sqrt{3} + \sqrt{3} - 8 \sqrt{3} \\ \\ 3 \sqrt{3} - 8 \sqrt{3} = - 5 \sqrt{3} \\ \\ f) \\ \\ \sqrt{300} + \sqrt{3} - \sqrt{27} = \sqrt{ {10}^{2} \times 3 } + \sqrt{3} - \sqrt{27} \\ \\ 10 \sqrt{3} + \sqrt{3} - \sqrt{27} \\ \\ 10 \sqrt{3} + \sqrt{3} - \sqrt{ {3}^{2} \times 3 } \\ \\ 10 \sqrt{3} + \sqrt{3} - 3 \sqrt{3} \\ \\ 11 \sqrt{3} - 3 \sqrt{3} = 8 \sqrt{3} \\ \\ g) \\ \\ \sqrt{150} : \sqrt{6} = \\ \\ \sqrt{150} \times 6 : 6 = \sqrt{900} : 6 = 30 : 6 = 5$