Bardzo pilne prosze pomóżcie!.8.8 wyznacz sumę, iloczyn i obie różnice przedziałów A i B gdy A=(-3:4.... Question from @spiner514

September 2018 1 20 Report

Bardzo pilne prosze pomóżcie!.
8.8 wyznacz sumę, iloczyn i obie różnice przedziałów A i B gdy
A=(-3:4), B=<1;7>

5.26 rozwiąż równanie :
a) (2x-1)(3-x)(x+4)=0

5.28 rozwiąż równanie:
d) (x-3)do potęgi 3 =243
h) x do potęgi 3 - x do potęgi 2=0

Roma

Zad. 8.8

$A = (-3; \ 4), \ B=\langle1; \ 7\rangle$

$A \cup B = (-3; \ 4) \cup \langle1; \ 7\rangle = (-3; \ 7 \rangle$

$A \cap B = (-3; \ 4) \cap \langle1; \ 7\rangle = \langle 1; \ 4)$

$A \backslash B = (-3; \ 4) \backslash \langle1; \ 7\rangle = (-3; \ 1)$

$B \backslash A = \langle1; \ 7\rangle \backslash (-3; \ 4) = \langle 4; \ 7 \rangle$

Zad. 5.26

$(2x-1)(3-x)(x+4)=0$

$2x-1 = 0 \vee 3-x = 0 \vee x+4=0$

$2x-1 = 0$

$2x=1 \ / : 2$

$x = \frac{1}{2}$

$3-x = 0$

$-x = - 3 \ / \cdot (-1)$

$x = 3$

$x+4=0$

$x = - 4$

Odp. $x = - 4 \ lub \ x = \frac{1}{2} \ lub \ x = 3$

Zad. 5.28

$(x-3)^3 = 243$

$(x-3)^3 - 243 = 0$

$(x-3)^3 - (\sqrt[3]{243})^3 = 0$

$[(x-3) - \sqrt[3]{243}][(x-3)^2 + (x-3) \cdot \sqrt[3]{243} + (\sqrt[3]{243})^2] = 0$

$(x-3 - \sqrt[3]{243})(x^2-6x + 9 + \sqrt[3]{243}x-3\sqrt[3]{243} + \sqrt[3]{243^2}) = 0$

$(x-3 - \sqrt[3]{3^5})(x^2-6x + \sqrt[3]{3^5}x+ 9-3\sqrt[3]{3^5} + \sqrt[3]{(3^5)^2}) = 0$

$(x-3 - 3^{\frac{5}{3}})(x^2-6x + 3^{\frac{5}{3}}x+ 9-3\cdot 3^{\frac{5}{3}}+ \sqrt[3]{3^{10}}) = 0$

$(x-3 - 3^{\frac{5}{3}})[x^2-(6 - 3^{\frac{5}{3}})x+ 9-3^{\frac{8}{3}}+ 3^{\frac{10}{3}}] = 0$

$x-3 - 3^{\frac{5}{3}} =0 \vee x^2-(6 - 3^{\frac{5}{3}})x+ 9-3^{\frac{8}{3}}+ 3^{\frac{10}{3}}= 0$

$x-3 - 3^{\frac{5}{3}} =0$

$x=3 + 3^{\frac{5}{3}}$

$x^2-(6 - 3^{\frac{5}{3}})x+ 9-3^{\frac{8}{3}}+ 3^{\frac{10}{3}}= 0$

$\Delta = [-(6 - 3^{\frac{5}{3}})]^2 - 4 \cdot 1 \cdot (9-3^{\frac{8}{3}}+ 3^{\frac{10}{3}}) = (3^{\frac{5}{3}} - 6)^2-36+4 \cdot 3^{\frac{8}{3}} - 4 \cdot 3^{\frac{10}{3}} =$

$= 3^{\frac{10}{3}} - 12 \cdot 3^{\frac{5}{3}} + 36 -36+4 \cdot 3^{\frac{8}{3}} - 4 \cdot 3^{\frac{10}{3}} = - 4 \cdot 3 \cdot 3^{\frac{5}{3}} +4 \cdot 3^{\frac{8}{3}} - 3 \cdot 3^{\frac{10}{3}} =$