A. 88
B. 99
C. 89
D. 87
C
\(\begin{array}{l} \frac{{x - 12}}{{77}} + \frac{{x - 11}}{{78}} = \frac{{x - 74}}{{15}} + \frac{{x - 73}}{{16}}\\ \Leftrightarrow \left( {\frac{{x - 12}}{{77}} - 1} \right) + \left( {\frac{{x - 11}}{{78}} - 1} \right) = \left( {\frac{{x - 74}}{{15}} - 1} \right) + \left( {\frac{{x - 73}}{{16}} - 1} \right)\\ \Leftrightarrow \left( {\frac{{x - 12 - 77}}{{77}}} \right) + \left( {\frac{{x - 11 - 78}}{{78}}} \right) = \left( {\frac{{x - 74 - 15}}{{15}}} \right) + \left( {\frac{{x - 73 - 16}}{{16}}} \right)\\ \Leftrightarrow \frac{{x - 89}}{{77}} + \frac{{x - 89}}{{78}} - \frac{{x - 89}}{{15}} - \frac{{x - 89}}{{16}} = 0 \Leftrightarrow \left( {x - 89} \right)\left( {\frac{1}{{77}} + \frac{1}{{78}} - \frac{1}{{15}} - \frac{1}{{16}}} \right) = 0 \end{array}\)
Nhận thấy \( \frac{1}{{77}} + \frac{1}{{78}} - \frac{1}{{15}} - \frac{1}{{16}} \ne 0\) nên \( \left( {x - 89} \right)\left( {\frac{1}{{77}} + \frac{1}{{78}} - \frac{1}{{15}} - \frac{1}{{16}}} \right) = 0 \Leftrightarrow x - 89 = 0 \Leftrightarrow x = 89\)
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