Question

Calcular e simplificar as operações .

a) (raiz quadrada de 49)+(raiz quadrada de 36)-(raiz quadrada de 25)=
b) 2 (raiz quadrada de 5) + 3(raiz quadrada de 5) - (raiz quadrada de 5 )=
c) 3 (raiz quadrada de 12) - (raiz quadrada de 147) + (raiz quadrada de 243)=
d) 3 (raiz quadrada de 7) - 2(raiz quadrada de 5) + 3(raiz quadrada de 5) - raiz quadrada de 7)=
e) 2 (raiz quadrada de 18) - 5(raiz quadrada de 8) - (raiz quadrada de 50 )=
f) (raiz quadrada de 18) + (raiz quadrada de 20 ) - 2(raiz quadrada de 2 ) - 3(raiz quadrada de 5)=

Answer 1

a)$\sqrt{49}+\sqrt{36}-\sqrt{25}=\\ =7+6-5=8$

b)$2\sqrt{5}+3\sqrt{5}-\sqrt{5}=\\ =(2+3-1)\sqrt{5}=4\sqrt{5}$

c)$3\sqrt{12}-\sqrt{147}+\sqrt{243}=\\ =3\sqrt{2^2\cdot3}-\sqrt{7^2\cdot3}+\sqrt{9^2\cdot3}=\\ =6\sqrt{3}-7\sqrt{3}+9\sqrt{3}=\\ =(6-7+9)\sqrt{3}=\\ =8\sqrt{3}$

d)$3\sqrt{7}-2\sqrt{5}+3\sqrt{5}-\sqrt{7}\\ =(-2+3)\sqrt{5}+(3-1)\sqrt{7}=\\ =\sqrt{5}+2\sqrt{7}$

e)$2\sqrt{18}-5\sqrt{8}-\sqrt{50}=\\= 2\sqrt{3^2\cdot2}-5\sqrt{2^2\cdot2}-\sqrt{5^2\cdot2}=\\ =6\sqrt{2}-10\sqrt{2}-5\sqrt{2}=\\ =(6-10-5)\sqrt{2}=\\ =-9\sqrt{2}$

f)$\sqrt{18}+\sqrt{20}-2\sqrt{2}-3\sqrt{5}=\\ =\sqrt{3^2\cdot2}-2\sqrt{2}+\sqrt{2^2\cdot5}-3\sqrt{5}=\\ =3\sqrt{2}-2\sqrt{2}+2\sqrt{5}-3\sqrt{5}=\\ =(3-2)\sqrt{2}+(2-3)\sqrt{5}=\\ =\sqrt{2}-\sqrt{5}$