5 uždavinys
Sprendimas.
| 3 |
| / (2-x) |
|
| 3 |
| / (2-x) |
Paaiškinimas: Keitimas $${\normalsize x}$$ = $${\normalsize \sqrt {2}}$$.
| 3 |
| / (2-saknis(2)) |
| 3* (2+saknis(2)) |
| / (2-saknis(2))/ (2+saknis(2)) |
$${\normalsize (2-\sqrt {2})\cdot (2+\sqrt {2})}$$ = $${\normalsize (2^{2}-\sqrt {2}^{2})}$$ Paaiškinimas:
Paaiškinimas: Pagal kvadratų skirtumo formulę
$${\normalsize (a+b)\cdot (a-b) = a^{2}-b^{2}}$$
Čia a = 2, b = saknis(2)
$${\normalsize (a+b)\cdot (a-b) = a^{2}-b^{2}}$$
Čia a = 2, b = saknis(2)
| 3* (2+saknis(2)) |
| / ( 2^2- saknis(2)^2) |
$${\normalsize \sqrt {2}^{2}}$$ = $${\normalsize 2}$$
| 3* (2+saknis(2)) |
| / ( 2^2-2) |
$${\normalsize 2^{2}}$$ = $${\normalsize 4}$$
| 3* (2+saknis(2)) |
| / (4-2) |
$${\normalsize 4-2}$$ = $${\normalsize 2}$$
| 3* (2+saknis(2)) |
| / (2) |
$${\normalsize (2)}$$ = $${\normalsize 2}$$
| 3* (2+saknis(2)) |
| / 2 |
$${\normalsize 3\cdot (2+\sqrt {2})}$$ = $${\normalsize 3\cdot 2+3\cdot \sqrt {2}}$$
| ( 3* 2+ 3* saknis(2)) |
| / 2 |
$${\normalsize 3\cdot 2}$$ = $${\normalsize 6}$$
| (6+ 3* saknis(2)) |
| / 2 |
$$\frac{3}{2-x}$$ = $$$$
$$\frac{3}{2-\sqrt {2}}$$ = $$$$
$$\frac{3\cdot (2+\sqrt {2})}{(2-\sqrt {2})\cdot (2+\sqrt {2})}$$ = $$$$
$$\frac{3\cdot (2+\sqrt {2})}{2^{2}-\sqrt {2}^{2}}$$ = $$$$
$$\frac{3\cdot (2+\sqrt {2})}{4-2}$$ = $$$$
$$\frac{3\cdot (2+\sqrt {2})}{2}$$ = $$$$
$$\frac{6+3\cdot \sqrt {2}}{2}$$ $$$$
Atsakymas: D