Cho hai biểu thức A = 2x + 6/ x- 5

a .Thay x = 1 (TMĐK) vào A, ta được:

\[A = \frac{{2.1 + 6}}{{1 - 5}} = \frac{8}{{ - 4}} =  - 2\]

Vậy \[A =  - 2\] khi x = 1.

b.  \[B = \left( {\frac{4}{{x - 5}} + \frac{{3x - 1}}{{{x^2} - 25}} - \frac{1}{{x + 5}}} \right).\frac{{{x^2} + 5x}}{{x + 4}}\]

 \[ = \left[ {\frac{{4\left( {x + 5} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} + \frac{{3x - 1}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} - \frac{{x - 5}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}} \right].\frac{{x\left( {x + 5} \right)}}{{x + 4}}\]

 \[ = \frac{{4\left( {x + 5} \right) + 3x - 1 - \left( {x - 5} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}.\frac{{x\left( {x + 5} \right)}}{{x + 4}}\]

 \[ = \frac{{6\left( {x + 4} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}.\frac{{x\left( {x + 5} \right)}}{{x + 4}}\]

 \[ = \frac{{6x}}{{x - 5}}\]

c. Cho \[P = \frac{A}{B}\]. Tìm x để \[P = \frac{3}{4}\].

 \[P = \frac{A}{B} = \frac{{2x + 6}}{{x - 5}}:\frac{{6x}}{{x - 5}} = \frac{{x + 3}}{{3x}}\] \[\left( {x \ne  - 4\,;\,\,x \ne  \pm 5\,;\,\,x \ne 0} \right)\]

Với \[P = \frac{3}{4}\] thì \[\frac{{x + 3}}{{3x}} = \frac{3}{4}\]

 Suy ra \[4\left( {x + 3} \right) = 3 \cdot 3x\]

\[\,4x + 12 = 9x\]

\[5x = 12\]

\[x = \frac{{12}}{5}\,\,\left( {{\rm{TM}}} \right)\]