równania wymierne

1985 · Post autor: **1985** » 4 maja 2010, o 23:46

\(\displaystyle{ \frac{5}{x-2} + \frac{3}{2} =0}\)

agulka1987 · Post autor: **agulka1987** » 4 maja 2010, o 23:55

\(\displaystyle{ \frac{5}{x-2}+ \frac{3}{2}=0}\)

\(\displaystyle{ D: x \in R \backslash \left[ 2\right]}\)

\(\displaystyle{ 10+3(x-2)=0}\)

\(\displaystyle{ 3x = -4}\)

\(\displaystyle{ x=- \frac{4}{3}}\)

1985 · Post autor: **1985** » 5 maja 2010, o 00:01

agulka1987 pisze:\(\displaystyle{ \frac{5}{x-2}+ \frac{3}{2}=0}\)

\(\displaystyle{ D: x \in R \backslash \left[ 2\right]}\)

\(\displaystyle{ 10+3(x-2)=0}\)

\(\displaystyle{ 3x = -4}\)

\(\displaystyle{ x=- \frac{4}{3}}\)

dziękuje

-- 5 maja 2010, o 00:02 --

frac{x}{x-5} = frac{x-2}{x-6} ???plis-- 5 maja 2010, o 00:05 --

agulka1987 pisze:\(\displaystyle{ \frac{5}{x-2}+ \frac{3}{2}=0}\)

\(\displaystyle{ D: x \in R \backslash \left[ 2\right]}\)

\(\displaystyle{ 10+3(x-2)=0}\)

\(\displaystyle{ 3x = -4}\)

\(\displaystyle{ x=- \frac{4}{3}}\)

dziękuje

agulka1987 · Post autor: **agulka1987** » 5 maja 2010, o 00:24

\(\displaystyle{ \frac{x}{x-5} = \frac{x-2}{x-6}}\)

\(\displaystyle{ x(x-6) = (x-2)(x-5)}\)

\(\displaystyle{ x^2-6x = x^2-5x-2x+10}\)

\(\displaystyle{ -6x+5x+2x = 10}\)

\(\displaystyle{ x=10}\)