Julkarynkiewicz
Rozwiązane

Potrzebuje pomocy z zadaniem z matmy na jutro daje naj
a) -(-1/4)^3+(1/4)^2
b)(5/6)^2-5^2/6+5/6^2
c)10*(-0.4)^3-0.4*10^3
d)8^2*1/8-(-1/8)^0*(-8)^2
Znak ^n to potega



Odpowiedź :

MertB

Cześć!

Szczegółowe wyjaśnienie:

a)

[tex] - ( - \frac{1}{4} ) {}^{3} + ( \frac{1}{4} ) {}^{2} = - ( - \frac{1}{64} + \frac{1}{16} ) = \frac{1}{64} + \frac{1}{16} = \frac{5}{64} [/tex]

b)

[tex]( \frac{5}{6} ) { }^{2} - \frac{5 {}^{2} }{6} + \frac{5}{6 {}^{2} } = \frac{25}{36} - \frac{25}{6} + \frac{5}{36} = - \frac{10}{3} [/tex]

c)

[tex]10 \cdot( - 0.4) {}^{3} - 0.4\cdot10 {}^{3} = \\ - 0.4 \cdot10(( - 0.4) {}^{2} + 10 {}^{2} ) \\ - \frac{2}{5} \cdot10(( - \frac{2}{5} ) { }^{2} + 100) \\ - 2 \cdot2( \frac{4}{25 } + 100) \\ - 2 \cdot2 \cdot \frac{2504}{25} \ = - \frac{10016}{25} [/tex]

d)

[tex]8 {}^{2} \cdot \frac{1}{8} - ( - \frac{1}{8} ) {}^{0} \cdot( - 8) {}^{2} = \frac{1}{8} \cdot8 {}^{2} - 1 \cdot8 {}^{2} = - 7 \cdot8 = - 56[/tex]

Basetla

[tex]a) \ -(-\frac{1}{4})^{3}+(\frac{1}{4})^{2} = -(-\frac{1}{64})+\frac{1}{16} = \frac{1}{64}+\frac{4}{64} = \frac{5}{64}[/tex]

[tex]b) \ (\frac{5}{6})^{2}-\frac{5^{2}}{6}+\frac{5}{6^{2}} = \frac{25}{36}-\frac{25}{6}+\frac{5}{36} = \frac{30}{36}-\frac{150}{36} = -\frac{120}{36} =-\frac{10}{3} = -3\frac{1}{3}[/tex]

[tex]c) \ 10\cdot(-0,4)^{3}-0,4\cdot10^{3} = 10\cdot(-0,064)-0,4\cdot1000 = -0,64-400 = -400,64[/tex]

[tex]d) \ 8^{2}\cdot\frac{1}{8}-(-\frac{1}{8})^{0}\cdot(-8)^{2} = 8-1\cdot64 = 8-64 = -56[/tex]

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