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pa solve po i really need it....
[tex] \frac2{r-1} + \frac4{r-5} =7[/tex]


Sagot :

[tex] \frac2{r-1} + \frac4{r-5} =7 \\\\r-1\neq 0 \ \ \wedge \ \ r-5\neq 0\\\\r \neq 1 \ \ \wedge \ \ r \neq 5 \\\\D=R\setminus \left \{ 1,5 \right \} \\\\ \frac {2(r-5)+4(r-1)} {(r-1)(r-5) }=7 [/tex]

[tex]\frac { 4r-10+4r-4} {(r-1)(r-5) }=7\ \ |\ common \ denominator \\\\\frac { 6r-14} {(r^2-5r-r+5 ) }=7\\\\7(r^2-6r +5)=6r-14 \\\\ 7r^2-42r +35=6r-14 =0 \ \ |\ subtract\ 6r \ and \ add \ 14 \ to\ both\ sides \\\\ 7r^2-42r +35-6r+14 =0 [/tex]

[tex]7r^2-48r +49 =0\\a=7, \ \ b=-48, \ \ c=49 \\ Quadratic \ Formula : \\ \\r_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{48-\sqrt{ (-48)^2-4 \cdot 7\cdot 49}}{2 \cdot 7}=\frac{48-\sqrt{ 2304- 1372}}{14}=\\\\=\frac{48-\sqrt{932}}{14}=\frac{48-\sqrt{4\cdot 233}}{14}=\frac{48-2\sqrt{233}}{14}=\frac{2(24- \sqrt{233})}{14}=\frac{ 24- \sqrt{233} }{7} \\\\r_{1}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{ 24+ \sqrt{233} }{7}[/tex]