Proszę o rozwiązanie zadań z załącznika.... Question from @r3fleX

kasia2517

30 14 17 5 4 17 1

----- · ------ - ----- · ------ ----- - ----- 4 - 4---- -¼

7 15 5 4 1 4 4

= ------------------------------------=------------------------=----------------=----------------- = -¼ · 34 = -8½

4 1 1 3 1 71 1 1

( 16 --- - 14-----) · ---- 2----- · ---- ---- · ----- ----

34 34 71 34 71 34 71 34

80 3 25 52

12 --- - 1---- - 10 ---- 1 ----

39 39 39 39 91 3 91

----------------------------- = ----------------- = ------- · ------ = --------- = 7

14 1 7 2 2 1 39 1 13

---- · ---- - -----·----- --- - -----

3 7 6 7 3 3

7¹²(3+4) 7¹² · 7

-------------- = -------------= 7

(7⁴)³ 7¹²

3¹⁴ (8·3-1) 3¹⁴ · 23 1

---------------=------------- = -----

46 · 3¹⁴ 46 · 3¹⁴ 2

8+3(5x²-25-x⁴+5x²)-(x⁴-16x²+64)+4x⁴-4x²+2x³-2x-x²+1+2x=

8 + 15x² - 75x - 3x⁴ + 15x² - x⁴ + 16x² - 64 + 4x⁴ - 4x² + 2x³ - 2x - x² + 1 + 2x =

2x³ + 41x² - 75x - 55

13-2(3x²-9-x⁴+3x²)+x⁴+10x²+25-(3x⁴-12x²-3x³+12x-x²+4)-3x³=

13 - 6x² + 18 + 2x⁴ - 6x² + x⁴ + 10x² + 25 - 3x⁴ + 12x² + 3x³ - 12x + x² - 4 - 3x³ =

11x² - 12x + 52

POWODZENIA

0 votes Thanks 1

nitra017

ZADANIE 1.

a) $\frac{\frac{30}{7} \cdot \frac{14}{15} - \frac{17}{5} \cdot \frac{5}{4} }{(16\frac{4}{34} - 14\frac{1}{34}) \cdot \frac{1}{71} } = \\ \\= \frac{4-\frac{17}{4}}{2\frac{3}{34} \cdot \frac{1}{71} }= \\ \\ = \frac{4-4\frac{1}{4}}{\frac{71}{34} \cdot \frac{1}{71} }=\\ \\ =\frac{-\frac{1}{4}}{ \frac{1}{34} }= \\ \\ = -\frac{1}{4} \cdot \frac{34}{1} = -\frac{34}{4}=-8\frac{2}{4}=-8\frac{1}{2}$

b) $\frac{14\frac{2}{39} - 1\frac{1}{13} - 10\frac{25}{39} }{4\frac{2}{3} : 7 + 1\frac{1}{6} \cdot ( -\frac{2}{7}) } = \\ \\= \frac{14\frac{2}{39} - 1\frac{3}{39} - 10\frac{25}{39} }{\frac{14}{3} \cdot \frac {1}{7} + \frac{7}{6} \cdot ( -\frac{2}{7}) } = \\ \\ = \frac{1\frac{13}{39} }{\frac{2}{3} - \frac{1}{3} } = \\ \\ = \frac{1\frac{1}{3} }{\frac{1}{3}}= \\ \\ = \frac{\frac{4}{3} }{\frac{1}{3}} =\\ \\ = \frac{4}{3} \cdot 3= \\ \\ =4$

c) $\frac{7^{12} \cdot 3+4 \cdot 7^{12} }{ (7^{12} : 7^{8})^{3}}= \frac{7^{12} \cdot (3+4) }{ (7^{12-8} )^{3}}= \\ \\ =\frac{7^{12} \cdot 7 }{ (7^{4} )^{3}} = \frac{7^{13} }{ 7^{12}}=7$

d) $\frac{8 \cdot 3^{4} \cdot 3^{11} - 3^{2} \cdot 3^{12}}{ 46 \cdot (3^{18} : 3^{4})}= \\ \\= \frac{8 \cdot 3 \cdot 3^{14} - 3^{14}}{ 46 \cdot (3^{18} : 3^{4})}=\\ \\ = \frac{8 \cdot 3 \cdot 3^{14} - 3^{14}}{ 46 \cdot 3^{14}}= \\ \\ = \frac{3^{14} \cdot (8 \cdot 3-1)}{ 46 \cdot 3^{14}} =\\ \\ = \frac{3^{14} \cdot 23}{ 46 \cdot 3^{14}} = \frac{1}{2}$

ZADANIE 2.

a) $8+3(5-x^{2})(x^{2}-5)-(x^{2}-8^{2})+ \\ (4x^{2}+2x-1)(x^{2}-1)+2x=$

$8+3(5x^{2}-25-x^{4}+5x^{2})-\\(x^{4}-16x^{2}+64)+4x^{4}\\-4x^{2}+2x^{3}-2x-x^{2}+1+2x=$

$8 + 15x^{2} - 75x - 3x^{4} + 15x^{2} \\- x^{4} + 16x^{2} - 64+ 4x^{4}\\+ 2x^{3} - 5x^{2}+1 =$

$2x^{3} + 41x^{2} - 75x - 55$

b) $13-2(3-x^{2})(x^{2}-3)+(x^{2}+5)^{2}\\-(3x^{2}-3x-1)-3x^{3}=$

$13-2(3x^{2}-9-x^{4}+3x^{2})+x^{4}+\\10x^{2}+25-(3x^{4}-12x^{2}-3x^{3}\\+12x-x^{2}+4)-3x^{3}=$

$13 - 6x^{2} + 18 + 2x^{4} - 6x^{2} \\+ x^{4} + 10x^{2} + 25 - 3x^{4} \\+ 12x^{2} + 3x^{3} - 12x + x^{2} - 4 - 3x^{3} =$

$11x^{2} - 12x + 52$

1 votes Thanks 1

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