przekształcenia algebraiczne i ich zastosowania
xxx: 1.Udowodnij że:jeśli ... to...
a) x+1/x=3 to x(do drugiej potegi) + 1/x (do 2 pot)
b)x−1/x=1 to x2potega + 1/ x2potegi =3
2.I kolejne zad dane są liczby x=3− pierwiastek z 5 i y=4+2 pierwiastki z 5
A) x*y
B) x/y
C)3y−x
D)x do 2 potęgi − y do drugiej potegi
3. usuń niewymierność z mianownika 2/ 1 − √2 + √3
1 lis 22:44
Eta:
| | 1 | | 1 | |
to: x2+ |
| =(x+ |
| )2−2= 9−2=7 |
| | x2 | | x | |
| | 1 | | 1 | |
to: x2+ |
| = (x− |
| )2+2= 1+2=3 |
| | x2 | | x | |
1 lis 22:48
5-latek:
1 lis 23:09
pigor: ..., np. tak:
| | 1 | | 1−√2−√3 | |
3) |
| = |
| = |
| | 1−√2+√3 | | (1−√2+√3)(1−√2−√3) | |
| | 1−√2−√3 | | 1−√2−√3 | |
= |
| = |
| = |
| | (1−√2)2−(√3)2 | | 1−2√2+2−3 | |
| | 1−√2−√3 | | (1−√2−√3)*√2 | | √2−2−√6 | |
= |
| = |
| = |
| = |
| | −2√2 | | −2√2*√2 | | −2*2 | |
| | √2−√6−2 | |
= |
| = − 14 (√2−√3−2)= 14 (2−√2+√3). ...  |
| | −4 | |
2 lis 01:25