Hejj, potrzebuje zadanie 11przyklady e,f
12 a,b,c,d
14a,b
Z gory dziekuje za pomoc

[tex]e) \ \frac{1}{3}m^{3}+\frac{5}{6}m^{3}-\frac{1}{12}m^{3} = \frac{4}{12}m^{3}+\frac{10}{12}m^{3}-\frac{1}{12}m^{3} =\frac{13}{12}m^{3} = 1\frac{1}{12}m^{3}\\\\f) \ 0,1w\nu^{2}-\frac{2}{5}w\nu^{2}-4w\nu^{2} = 0,1w\nu^{2} - 0,4w\nu^{2}-4w\nu^{2} = -4,3w\nu^{2}[/tex]

12.

[tex]a) \ -x-4(1-x) = -x-4+4x = 3x-4\\\\b) \ 3(5x-2)-5(3-4x) = 15x-6-15+20x = 35x-21\\\\c) \ 5(3-x)-2(1-5x) = 15-5x-2+10x = 5x+13\\\\d) \ 3x(5-2x)+5(4x-3) = 15x - 6x^{2}+20x-15 =-6x^{2}+35x-15[/tex]

14.

[tex]a) \ (4m - 2n)(3m-5n) = 12m^{2}-20mn-6mn+10n^{2} = 12m^{2}-26mn+10n^{2}\\\\b) \ (3+2b)(1-b+5c) = 3 - 3b+15c+2b-2b^{2}+10bc = -2b^{2}-b+15c+10bc+3[/tex]

Wyjaśnienie:

Jeżeli przed nawiasem mamy znak minus (-), to po uwolnieniu nawiasu zmieniamy znaki na przeciwne.

(a + b)(c + d) = ac + ad + bc + bd

(a + b)(c + d + e) = ac + ad + ae + bc + cd + be

ZbiorJviz ZbiorJviz · Answer 2

Odpowiedź:

[tex]\huge\boxed{ zad. 11 }[/tex]

[tex]\boxed{e)~~\dfrac{1}{3} m^{3}+\dfrac{5}{6} m^{3}-\dfrac{1}{12} m^{3} =1\dfrac{1}{12}\right) m^{3}}[/tex]

[tex]\boxed{f)~~0,1wv^{2}-\dfrac{2}{5} wv^{2}-4wv^{2}=-4,3wv^{2}}[/tex]

[tex]\huge\boxed{ zad. 12 }[/tex]

[tex]\boxed{a)~~-x-4\cdot (1-x)=3x-4}[/tex]

[tex]\boxed{b)~~3\cdot (5x-2)-5\cdot (3-4x)=35x-21}[/tex]

[tex]\boxed{c)~~5\cdot (3-x)-2\cdot (1-5x)=5x+13}[/tex]

[tex]\boxed{d)~~3x\cdot (5-2x)+5\cdot (4x-3)=-6x^{2} +35x-15}[/tex]

[tex]\huge\boxed{ zad. 14 }[/tex]

[tex]\boxed{a)~~(4m-2n)\cdot (3m-5n)=12m^{2} -26mn+10n^{2} }[/tex]

[tex]\boxed{b)~~(3+2b)\cdot (1-b+5c)=-2b^{2}-b +15c+10bc+3}[/tex]

Szczegółowe wyjaśnienie:

Dla przypomnienia:

jednomianem ⇒ wyrażenie algebraiczne będące iloczynem liczby oraz zmiennych.
Liczbę stojącą przy zmiennej nazywa się współczynnikiem jednomianu.
[tex](a+b)\cdot (c-d)=a\cdot c -a\cdot d + b\cdot c -b\cdot d[/tex]
[tex]-(a+b-c)=-a-b+c[/tex]
[tex](-)\cdot (-)=(+)[/tex]
[tex](-)\cdot (+)=(-)[/tex]
[tex]x^{n} \cdot x^{m} =x^{n+m}[/tex]

Rozwiązujemy:

[tex]\huge\boxed{ zad. 11 }[/tex]

[tex]\boxed{e)}~~\dfrac{1}{3} m^{3}+\dfrac{5}{6} m^{3}-\dfrac{1}{12} m^{3} =\left(\dfrac{1}{3} +\dfrac{5}{6} -\dfrac{1}{12}\right) m^{3}=\left(\dfrac{1\cdot 4}{3\cdot 4} +\dfrac{5\cdot 2}{6\cdot 2} -\dfrac{1}{12}\right) m^{3}=\left(\dfrac{4}{12} +\dfrac{10}{12} -\dfrac{1}{12}\right) m^{3}=\left(\dfrac{14}{12} -\dfrac{1}{12}\right) m^{3}=\dfrac{13}{12}\right) m^{3}=\boxed{1\dfrac{1}{12}\right) m^{3}}[/tex]

[tex]\boxed{f)}~~0,1wv^{2}-\dfrac{2}{5} wv^{2}-4wv^{2}=\dfrac{1}{10} wv^{2}-\dfrac{2}{5} wv^{2}-4wv^{2}=\left(\dfrac{1}{10} -\dfrac{2\cdot 2}{5\cdot 2} -4\right)wv^{2}=\left(\dfrac{1}{10} -\dfrac{4}{10} -4\right)wv^{2}=\left(-\dfrac{3}{10} -4\right)wv^{2}=-4\dfrac{3}{10} wv^{2}=\boxed{-4,3wv^{2}}[/tex]

[tex]\huge\boxed{ zad. 12 }[/tex]

[tex]\boxed{a)}~~-x-4\cdot (1-x)=-x-4\cdot 1-4\cdot (-x)=-x-4+4x=\boxed{3x-4}[/tex]

[tex]\boxed{b)}~~3\cdot (5x-2)-5\cdot (3-4x)=3\cdot 5x+3\cdot (-2)-5\cdot 3-5\cdot (-4x)=15x-6-15+20x=15x+20x-6-15=\boxed{35x-21}[/tex]

[tex]\boxed{c)}~~5\cdot (3-x)-2\cdot (1-5x)=5\cdot 3+5\cdot (-x)-2\cdot 1-2\cdot (-5x)=15-5x-2+10x=-5x+10x+15-2=\boxed{5x+13}[/tex]

[tex]\boxed{d)}~~3x\cdot (5-2x)+5\cdot (4x-3)=3x\cdot 5+3x\cdot (-2x)+5\cdot 4x+5\cdot (-3)=15x-6x^{2} +20x-15=\boxed{-6x^{2} +35x-15}[/tex]

[tex]\huge\boxed{ zad. 14 }[/tex]

[tex]\boxed{a)}~~(4m-2n)\cdot (3m-5n)=4m\cdot 3m+4m\cdot (-5n)-2n\cdot 3m-2n\cdot (-5n)=12m^{2} -20mn-6mn+10n^{2} =\boxed{12m^{2} -26mn+10n^{2} }[/tex]

[tex]\boxed{b)}~~(3+2b)\cdot (1-b+5c)=3\cdot 1+3\cdot (-b)+3\cdot 5c+2b\cdot 1 +2b\cdot (-b)+2b\cdot 5c=3-3b+15c+2b-2b^{2} +10bc=\boxed{-2b^{2} -b+15c+10bc+3}[/tex]