1. Rozwiąż równości i nierówności

[MadeleineSwiecicka]

a) \(\displaystyle{ \frac{7}{x-5}= \frac{3}{x-3}}\)

b)\(\displaystyle{ \frac {x^2+7x+12} {x^2-1} = 0}\)

c) \(\displaystyle{ \frac {9x^2 - 5x} {3x-2}= 2 +\frac {x} {3x-2}}\)

d) \(\displaystyle{ \frac {9} {12x-1}=\frac {3} {4x+3}}\)

e) \(\displaystyle{ x^{2} - x - 30 = 0}\)

f) \(\displaystyle{ x^{2}+2 \frac {2}{3}x - 1 =0}\)

g) \(\displaystyle{ (3x-2)^{2}-x^{2} - 2x=1}\)

h)\(\displaystyle{ \frac {1}{3}- 4 \frac {1}{2}x+15=0}\)

i) \(\displaystyle{ x(x+19)\le3(18+5x)}\)

j) \(\displaystyle{ 5(x+1) < x (3-x)}\)

k) \(\displaystyle{ 8x + 8 > 2x^{2} + 8x - 6}\)

Proszę o pomoc.

[binio]

a) \(\displaystyle{ \frac{7}{x-5}= \frac{3}{x-3}}\)

\(\displaystyle{ x\neq 3 \vee x \neq 5}\)

\(\displaystyle{ 3x-15 = 7x-21}\)
\(\displaystyle{ 4x = 6}\)
\(\displaystyle{ x = \frac{3}{2}}\)

-- 25 sty 2016, o 21:44 --

b)\(\displaystyle{ \frac {x^2+7x+12} {x^2-1} = 0}\)

\(\displaystyle{ x \neq 1 \vee x \neq -1}\)

\(\displaystyle{ x^2+7x+12 = 0}\)

\(\displaystyle{ \Delta = 49 - 48 = 1}\)
\(\displaystyle{ x_{1} = \frac{-7-1}{2} = -4}\)
\(\displaystyle{ x_{2} = \frac{-7+1}{2} = -3}\)

-- 25 sty 2016, o 21:51 --

c) \(\displaystyle{ \frac {9x^2 - 5x} {3x-2}= 2 +\frac {x} {3x-2}}\)

\(\displaystyle{ 3x-2 \neq 0}\)
\(\displaystyle{ 3x \neq 2}\)
\(\displaystyle{ x \neq \frac{2}{3}}\)

\(\displaystyle{ 9x^2-5x = 6x-4 + x}\)
\(\displaystyle{ 9x^2-12x+4 = 0}\)

\(\displaystyle{ \Delta = 144 - 144 = 0}\)

\(\displaystyle{ x = \frac{12}{18} = \frac{2}{3}}\)

Brak rozwiazan

-- 25 sty 2016, o 21:59 --

d) \(\displaystyle{ \frac {9} {12x-1}=\frac {3} {4x+3}}\)

\(\displaystyle{ 4x+3 \neq 0}\)
\(\displaystyle{ 4x \neq -3}\)
\(\displaystyle{ x \neq -\frac{3}{4} \vee x \neq \frac{1}{12}}\)

\(\displaystyle{ 36x+27 = 36x-3}\)
\(\displaystyle{ 30 = 0}\)

Brak rozwiazan

-- 25 sty 2016, o 22:10 --

e) \(\displaystyle{ x^{2} - x - 30 = 0}\)

\(\displaystyle{ \Delta = 1 + 120 = 121}\)
\(\displaystyle{ x_{1} = \frac{1-11}{2} = -5}\)
\(\displaystyle{ x_{2} = \frac{1+11}{2} = 6}\)

-- 25 sty 2016, o 22:15 --

f) \(\displaystyle{ x^{2}+2 \frac {2}{3}x - 1 =0}\)

\(\displaystyle{ x^2+\frac{8}{3}x - 1 = 0}\)
\(\displaystyle{ 3x^2+8x-3 = 0}\)

\(\displaystyle{ \Delta = 64 + 36 = 100}\)
\(\displaystyle{ x_{1} = \frac{-8-10}{6} = -3}\)
\(\displaystyle{ x_{2} = \frac{-8+10}{6} = \frac{1}{3}}\)

-- 25 sty 2016, o 22:22 --

g) \(\displaystyle{ (3x-2)^{2}-x^{2} - 2x=1}\)

\(\displaystyle{ 9x^2-12x+4-x^2-2x = 1}\)
\(\displaystyle{ 8x^2-14x+3 = 0}\)

\(\displaystyle{ \Delta = 196 - 96 = 100}\)

\(\displaystyle{ x_{1} = \frac{14-10}{16} = \frac{1}{4}}\)
\(\displaystyle{ x_{2} = \frac{14+10}{16} = \frac{3}{2}}\)

-- 25 sty 2016, o 22:26 --

h)\(\displaystyle{ \frac {1}{3}- 4 \frac {1}{2}x+15=0}\)

Chyba zabraklo Ci tu \(\displaystyle{ x^{2}}\)

[MadeleineSwiecicka]

Chyba zabraklo Ci tu \(\displaystyle{ x^{2}}\)

Tak! Zabrakło! dzięki za czujność:)

[binio]

i) \(\displaystyle{ x(x+19)\le3(18+5x)}\)

\(\displaystyle{ x^2+19x \le 54+15x}\)
\(\displaystyle{ x^2+4x-54 \le 0}\)

\(\displaystyle{ \Delta = 16 + 216 = 232}\)

\(\displaystyle{ x_{1} = \frac{-4-2\sqrt{58}}{2} = -2-\sqrt{58}}\)
\(\displaystyle{ x_{2} = -2+\sqrt{58}}\)

\(\displaystyle{ x\in[-2-\sqrt{58}; -2+\sqrt{58}]}\)

-- 25 sty 2016, o 22:38 --

j) \(\displaystyle{ 5(x+1) < x (3-x)}\)

\(\displaystyle{ 5x+5 < 3x-x^2}\)
\(\displaystyle{ x^2+2x+5 < 0}\)

\(\displaystyle{ \Delta = 4 - 20 = -16}\)

\(\displaystyle{ x \in \emptyset}\)

-- 25 sty 2016, o 22:39 --

-- 25 sty 2016, o 22:48 --

k) \(\displaystyle{ 8x + 8 > 2x^{2} + 8x - 6}\)

Proszę o pomoc.

\(\displaystyle{ 2x^2-14 < 0}\)

\(\displaystyle{ x = \sqrt{7}}\) lub \(\displaystyle{ x = -\sqrt{7}}\)

\(\displaystyle{ x\in(-\sqrt{7}; \sqrt{7})}\)