Phương trình \(\frac{8}{{x - 8}} + \frac{{11}}{{x - 11}} = \frac{9}{{x - 9}} + \frac{{10}}{{x - 10}}\) có nghiệm là

Đáp án đúng là: C

Điều kiện xác định: x ≠ 8, x ≠ 9, x ≠ 10, x ≠ 11.

Ta có: \(\frac{8}{{x - 8}} + \frac{{11}}{{x - 11}} = \frac{9}{{x - 9}} + \frac{{10}}{{x - 10}}\)

\(\frac{8}{{x - 8}} + 1 + \frac{{11}}{{x - 11}} + 1 = \frac{9}{{x - 9}} + 1 + \frac{{10}}{{x - 10}} + 1\)

\(\frac{x}{{x - 8}} + \frac{x}{{x - 11}} = \frac{x}{{x - 9}} + \frac{x}{{x - 10}}\)

\(\frac{x}{{x - 8}} + \frac{x}{{x - 11}} - \frac{x}{{x - 9}} - \frac{x}{{x - 10}} = 0\)

\(x\left( {\frac{1}{{x - 8}} + \frac{1}{{x - 11}} - \frac{1}{{x - 9}} - \frac{1}{{x - 10}}} \right) = 0\)

TH1: x = 0 (thỏa mãn)

TH2: \(\frac{1}{{x - 8}} + \frac{1}{{x - 11}} - \frac{1}{{x - 9}} - \frac{1}{{x - 10}} = 0\)

\(\frac{1}{{x - 8}} + \frac{1}{{x - 11}} = \frac{1}{{x - 9}} + \frac{1}{{x - 10}}\)

\(\frac{{x - 11 + x - 8}}{{\left( {x - 8} \right)\left( {x - 11} \right)}} = \frac{{x - 10 + x - 9}}{{\left( {x - 9} \right)\left( {x - 10} \right)}}\)

\(\frac{{2x - 19}}{{\left( {x - 8} \right)\left( {x - 11} \right)}} = \frac{{2x - 19}}{{\left( {x - 9} \right)\left( {x - 10} \right)}}\)

\(\left( {2x - 19} \right)\left( {x - 8} \right)\left( {x - 11} \right) = \left( {2x - 19} \right)\left( {x - 9} \right)\left( {x - 10} \right)\)

(2x – 19)(x² – 19x + 88) – (2x – 19)(x² – 19x + 90) = 0

(2x – 19)(x² – 19x + 88 – x² + 19x – 90) = 0

−2(2x – 19) = 0 khi 2x – 19 = 0 hay x = \(\frac{{19}}{2}\) (thỏa mãn).

Vậy nghiệm của phương trinh là x = \(\frac{{19}}{2}\) hoặc x = 0.