Rozwiązane

pomoże jakaś dobra dusza?​

Pomoże Jakaś Dobra Dusza class=

Odpowiedź :

Cześć ;-)

a)

[tex](3-\sqrt2)(\sqrt2-1)=\\\\=3\sqrt2-3-\sqrt{2^2}+\sqrt2=\\\\=3\sqrt2-3-2+\sqrt2=\boxed{4\sqrt2-5}[/tex]

b)

[tex]\sqrt3-(\sqrt3-5)(4+\sqrt3)=\\\\=\sqrt3-(4\sqrt3+\sqrt{3^2}-20-5\sqrt3)=\\\\=\sqrt3-(-\sqrt3+3-20)=\sqrt3-(-\sqrt3-17)=\\\\=\sqrt3+\sqrt3+17=\boxed{2\sqrt3+17}[/tex]

c)

[tex](\sqrt5-1)(\sqrt5+1)=\\\\=\sqrt{5^2}+\sqrt5-\sqrt5-1=\\\\=5+\sqrt5-\sqrt5-1=\boxed4[/tex]

d)

[tex](\sqrt3-2)(5+\sqrt2)-\sqrt6=\\\\=5\sqrt3+\sqrt6-10-2\sqrt2-\sqrt6=\boxed{5\sqrt3-2\sqrt2-10}[/tex]

Pozdrawiam!

123bodzioviz

Odpowiedź:

zad 4

a)

(3 - √2)(√2 - 1) = 3√2 - √2 * √2 - 3 + √2 = 3√2 - 2 - 3 + √2 = 4√2 - 5

b)

√3 - (√3 - 5)(4 + √3) = √3 - (4√3 - 20 + √3 * √3 - 5√3) = √3 - (- √3 - 20 + 3) =

= √3 + √3 + 20 - 3 = 2√3 + 17

c)

(√5 - 1)(√5 + 1) = (√5)² - 1 = 5 - 1 = 4

d)

(√3 - 2)(5 + √2) - √6 = 5√3 - 10 + √3 * √2 - 2√2 - √6 =

= 5√3 - 10 + √6 - 2√2 - √6 = 5√3 - 2√2 - 10

Viz Inne Pytanie