Question

1. OBLICZ
64⅓+64½+64⁻¹=
27⅓-3²+9⁻¹=
0,01½-10⁻¹-4⁰=
2½-4½+8⅓=
(7 123/511)⁰+0⁵-(-1⁷)=
(½)³+(1½)⁻¹-(2½)⁰=
2. OBLICZ
4√12+2√75-3√27=
2√50-√98+2√200=
(2√3+√6)²=
3. PODAJ PPRZYBLIZENIE Z DOKLADNOSCIA DO 0,01 LICZBY:
√2; 7/11; √3-√2.

Answer 1

64⅓+64½+64⁻¹=128,85
27⅓-3²+9⁻¹=18,44
0,01½-10⁻¹-4⁰=-0,59
(7 123/511)⁰+0⁵-(-1⁷)=2
(½)³+(1½)⁻¹-(2½)⁰=1,79
4√12+2√75-3√27=15,59
2√50-√98+2√200=23pierwiastek z 2 czyli 32,52
(2√3+√6)²=18+12pod pierw.2 czyli 34,97
√2; 7/11; √3-√2. .... 1,41;0,64;0,32

Answer 2

Zad. 1
a) 64⅓+64½+64⁻¹=2⁶x⅓+2⁶+½+2⁻⁶=2⁶x(⅓+½+2⁻¹)=2⁶(2/6+3/6+3/6)=2₆x8/6=64x8/6=32x8/3=256/3=85⅓
b) 27⅕-3²+9⁻¹=3³x3⁻¹-3²+3²=3²-3²+3⁻²=3⁻²=1/9
c) 0.01½-10⁻¹-4⁰=0.01x0,5-1/10-1=0,005-0,1-1=-0,095-1=-1,095
d) 2½-4½+8⅓=3/2-9/2+25/3=-6/2+25/3=-18/6+50/6=32/6=5 2/6=5⅓
e)(7 123/511)⁰+0⁵-(-1⁷)=1+1+1=3
f) (½)³+(1½)⁻¹-(2½)⁰=1/8+(3/2)⁻¹-1=1/8+2/3-1=3/24+16/24-1=19/24-1=-5/24
Zad. 2
a)4√12+2√75-3√27=4√4x3+2√3x25-3√9x3=4x2√3+2x5√3-3x3√3=8√3+10√3-9√3=9√3
b)2√50-√98+2√200=2√2x25-√2x49+2√2x100=2x5√2-7√2+2x10√2=10√2+7√2-20√2=-3√2
c)(2√3+√6)²=(2√3)²+(√6)²=4x3+6=12+6=18
Zad.3
a)√2≠1,41
b)7/11≈0,64
c)√3-√2≈1,73-1,41=0,32

Answer 3

1. OBLICZ 64⅓+64½+64⁻¹=4+8+1/64=12i1/64 27⅓-3²+9⁻¹=3-9+1/9=-5i8/9 0,01½-10⁻¹-4⁰=1/10-1/10-1=-1 2½-4½+8⅓=pierwiastek z 2-2+2=pierwiastek z 2 (7 123/511)⁰+0⁵-(-1⁷)=1+0-(-1)=2 (½)³+(1½)⁻¹-(2½)⁰=1/8+2/3-1=19/24-1=-5/24 2. OBLICZ 4√12+2√75-3√27= 2√50-√98+2√200= (2√3+√6)²= 3. PODAJ PPRZYBLIZENIE Z DOKLADNOSCIA DO 0,01 LICZBY: √2; 7/11; √3-√2.