Răspuns:
1.
[tex]4 \sqrt{27} - 5 \sqrt{12} + 3 \sqrt{48} - 2 \sqrt{75} = \\ 12 \sqrt{ 3} - 10 \sqrt{3} + 12 \sqrt{3} - 10 \sqrt{3} = \\ (12 - 10 + 12 - 10) \sqrt{3} = \\ 4 \sqrt{3} [/tex]
2.
[tex]0.(3) {x}^{2} = \frac{27}{256} \\ \frac{3}{9} {x}^{2} = \frac{27}{256} \\ \frac{1}{3} {x}^{2} = \frac{27}{256} \: \: \: | \: \times 3 \\ {x}^{2} = \frac{81}{256} \\ x = + - \sqrt{ \frac{81}{256} } \\ x = + - \frac{9}{16} [/tex]
3.
| 2x-5 | =9
-> 2x-5 =9 => 2x=14 => x=7
-> 2x-5 = -9 => 2x= -4 => x= -2
4.
ip=20
c1=12
=> prin teorema lui Pitagora
c2²= ip²-c1²
c2²= 20²-12² = 400- 144 = 256
c2= radical din 256 = 16
perimetrul p = ip+c1+c2 = 20+12+16 = 32+16 =48
Ex.5 si 6 nu stiu sa le fac ....scuze
Sper ca te-am ajutat :)
