| 1 | √3 | 1 | ||||
Oblicz dokładną wartość wyrażenia | − | + | sin30 | |||
| sin10 | cos10 | 2 |
| 1 | ||
2x2*0,52x+0,5> | (przy pierwszej dwojce jest 2 do potegi x a ten x do potegi 2 , | |
| √8 |
| 3x+3−x | 3 | 2 | 1−4x | ||||
=2 , | + | = 0 , | =2x+1 | ||||
| 3x−3−x | 2x−1 | 2−2x | 1−2x |
| 1 | 1 | |||
Otoz 3√(a+1)−3=3√ | = | |||
| (a+b)3 | a+b |
| 1 | ||
czy moge tez napisac tak 3√(a+1)−3= 3√[(a+1)−3]−1=(a+b)−1= | ? | |
| a+b |
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Witam. Wykazać, że | + | + | + | + | = | |||||||||||||||||||||||||||||||||||||||||||
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Wykorzystaj wzór: | + | = | ||||||||||||||||||||||
| 1 | 1 | (1/9)−1/2 a−2b−1 | ||||
( | + | ): | ||||
| b−√a | b+√a | a−2−a−1b−2 |
| 3√2−1 * 3√2+1 | 1 | 1 | |||
≤ ( | )x2−x ≤ 6π * 31−π * ( | )π−1 | |||
| 3√8 | 6 | 2 |
| (1+a)3√1+a | ||
√ | * U{√3{9+18a−1+9a−2} i to drugie wyrazenie ma byc pod | |
| 3a |
| 1 | a2+3a+2 | ||
a*4√(a+1)(a2−1)(1+2a+a2)*( | )−1= | ||
| 2 | √a−1 |
| 1 | (a+1)(a+2) | ||
a*4√(a+1)(a+1)(a−1)(a+1)2*( | )−1= | ||
| 2 | √a−1 |
| 1 | √a−1 | ||
a*4√(a+1)4(a−1)* | = | ||
| 2 | (a+1)(a+2) |
| 1 | |a+1|4√(a−1)3 | ||
a* | |||
| 2 | (a+1)(a+2) |
| a | a | |||
a) 1+sina=(sin | +cos | )2 | ||
| 2 | 2 |
| a | a | |||
b)cosa=cos2 | −sin2 | |||
| 2 | 2 |
| sin22a | ||
c)sin4a+cos4a=1− | ||
| 2 |
| dx | ||||||||
∫ | |||||||||
| x(x+1)2 |
| 1 | 1 | 1 | ||||
podstawiając mam: ∫ | − | − | ||||
| x | x+1 | (x+1)2 |
| cos270+2cos(−60) | sin(180+90)−2cos60 | −1−1 | |||
= | = | = 2 | |||
| tg(−225) | −tg(180+45) | −1 |
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