DAJE NAJJJ!!!?
MATEMATYKA KLASA 7!​


DAJE NAJJJMATEMATYKA KLASA 7 class=

Odpowiedź :

POZIOM A

a) 4k - 8        dla  k = 2

4 * 2 - 8 = 8 - 8 = 0

b) [tex]\frac{2}{3b-2}[/tex]         dla b = 3

[tex]\frac{2}{3*3 - 2} = \frac{2}{9 - 2} = \frac{2}{7}[/tex]

c) [tex]\frac{a^2 +5a}{6}[/tex]        dla a = 3

[tex]\frac{3^2 +5*3}{6} = \frac{9 +15}{6} = \frac{24}{6} = 4[/tex]

d) 2x + [tex]x^2[/tex]       dla  x = 4

2*4 + [tex]4^2[/tex] = 8 + 16 = 24

e) [tex]p^2 - p^3[/tex]          dla p = 2

[tex]2^2 - 2^3 = 4 - 8 = -4[/tex]

f) 3m - 1         dla m = 4

3*4 - 1 = 12 - 1 = 11

g) [tex]\frac{2}{5}(y+3)^2[/tex]        dla y = 7

[tex]\frac{2}{5} (7+3)^2 = \frac{2}{5} * 10^2 = \frac{2}{5} * 100 = \frac{200}{5} = 40[/tex]

h) t(t + 2)       dla t = 5

5(5+2) = 5 * 7  = 35

i) [tex]\frac{3z - 2z}{153}[/tex]        dla z = 0

[tex]\frac{3*0 - 2*0 }{153} = 0[/tex]

POZIOM B

a) 5m + [tex]m^2[/tex]   dla m = -3

5*(-3) + [tex](-3)^2[/tex] = 5*(-3) + 9 = -15 + 9 = -6

b) [tex]\frac{4,5 - 2b^2}{3}[/tex]       dla      b = -3

[tex]\frac{4,5 - 2*(-3)^2}{3} = \frac{4,5 - 2*9}{3} = \frac{4,5 - 18}{3} = \frac{-13,5}{3} = -4,5[/tex]

c) [tex]k^2 - k^3[/tex]          dla   k = -2

[tex](-2)^2 - (-2)^3 = 4 - (-8) = 4 + 8 = 12[/tex]

d) [tex]3x^2 - 2x[/tex]       dla  x = -1

[tex]3*(-1)^2 + 2*(-1) = 3*1 + (-2) = 3 - 2 = 1[/tex]

e) [tex]\frac{-7}{8-4a}[/tex]   dla a = -1

[tex]\frac{-7}{ 8-4*(-1)} = \frac{-7}{8 - (-4) } = \frac{-7}{8+4} = \frac{-7}{12} = - \frac{7}{12}[/tex]

f) 2(4p + 1)-5   dla p = -1

2(4*(-1) +1) - 5 = 2( -4 + 1) - 5 = 2 * (-3) - 5 = -6 - 5 = -11

g) y(y +3)   dla y = -4

-4(-4+3) = -4 * (-1) = 4

h) [tex]\frac{3}{4}(t-2)^2[/tex]      dla t = -6

[tex]\frac{3}{4}(-6-2)^2 = \frac{3}{4} * (-8)^2 = \frac{3}{4} * 64 = \frac{3*64}{4} = \frac{192}{4} = 48[/tex]

i) [tex]\frac{4z -2}{11}[/tex]  dla z = -2

[tex]\frac{4*(-2) - 2 }{11} = \frac{-8 - 2}{11} = \frac{-10}{11} = - \frac{10}{11}[/tex]