Question

Exercitiu 5 (problema cu radicali)​

Answer 1

Răspuns:

a) (√7+√3)(√7-√3) = √7^2 -√3^2 =7-3=4

b) (√5-√2)^2+2√2(√5-√2) =(√5)^2 -2*√5*√2+(√2)^2 +2√2*√5 +2√2 *(-√2) = 5 -2√10 +2 +2√10 +2 *(-2) =7-4 =3

c) (3+2√2)^2 -(3-√8)^2 =3^2 +2*3*2√2+(2√2)^2 -(3^2 -2*3*√8+√8^2)=9+12√2+4*2-(9-6*2√2+8) =9+12√2+8-(17-12√2) =17+12√2-17+12√2 =24√2

d) (√2-√3)^2 -(√2-√3)(√2+√3) = (√2)^2 -2*√2*√3+(√3)^2 -((√2)^2 -(√3)^2)

=2-2√6+3-(2-3)=2-2√6+3- (-1) =2+3+1-2√6= 6-2√6

Explicație pas cu pas:

Answer 2

Răspuns:

a) =

$\sqrt{7} ^{2} - \sqrt{21} + \sqrt{21} - \sqrt{3} ^{2}$

$7 - 3$

$= 4$

b)=

$\sqrt{5} ^{2} - 2 \times \sqrt{5} \times \sqrt{2} + 2 \sqrt{10} - 4$

$5 - 2 \sqrt{10 } + 2 \sqrt{10} - 4$

$5 - 4 = 1$

(pentru ca cele 2 numere se reduc raman doar 5-4 )