Google
Câu hỏi: Tính tổng $S={{a}^{{{\left( {{\log }_{2}}7 \right)}^{2}}}}+{{b}^{{{\left( {{\log }_{7}}9 \right)}^{2}}}}+{{c}^{{{\left( {{\log }_{9}}11 \right)}^{2}}}}$, biết rằng ${{a}^{{{\log }_{2}}7}}=4,{{b}^{{{\log }_{7}}9}}=49,{{c}^{{{\log }_{9}}11}}=81.$
A. $S=134.$
B. $S=8978.$
C. $S=251.$
D. $S=469.$
A. $S=134.$
B. $S=8978.$
C. $S=251.$
D. $S=469.$
Chú ý: Với điều kiện có nghĩa ta luôn có ${{a}^{{{\log }_{b}}c}}={{c}^{{{\log }_{b}}a}}.$ Khi đó ${{a}^{{{\log }_{a}}b}}=b.$
Ta có $S={{a}^{{{\left( {{\log }_{2}}7 \right)}^{2}}}}+{{b}^{{{\left( {{\log }_{7}}9 \right)}^{2}}}}+{{c}^{{{\left( {{\log }_{9}}11 \right)}^{2}}}}={{\left( {{a}^{{{\log }_{2}}7}} \right)}^{{{\log }_{2}}7}}+{{\left( {{b}^{{{\log }_{7}}9}} \right)}^{{{\log }_{7}}9}}+{{\left( {{c}^{{{\log }_{9}}11}} \right)}^{{{\log }_{9}}11}}$
$={{4}^{{{\log }_{2}}7}}+{{49}^{{{\log }_{7}}9}}+{{81}^{{{\log }_{9}}11}}={{\left( {{2}^{{{\log }_{2}}7}} \right)}^{2}}+{{\left( {{7}^{{{\log }_{7}}9}} \right)}^{2}}+{{\left( {{9}^{{{\log }_{9}}11}} \right)}^{2}}={{7}^{2}}+{{9}^{2}}+{{11}^{2}}=251.$
Ta có $S={{a}^{{{\left( {{\log }_{2}}7 \right)}^{2}}}}+{{b}^{{{\left( {{\log }_{7}}9 \right)}^{2}}}}+{{c}^{{{\left( {{\log }_{9}}11 \right)}^{2}}}}={{\left( {{a}^{{{\log }_{2}}7}} \right)}^{{{\log }_{2}}7}}+{{\left( {{b}^{{{\log }_{7}}9}} \right)}^{{{\log }_{7}}9}}+{{\left( {{c}^{{{\log }_{9}}11}} \right)}^{{{\log }_{9}}11}}$
$={{4}^{{{\log }_{2}}7}}+{{49}^{{{\log }_{7}}9}}+{{81}^{{{\log }_{9}}11}}={{\left( {{2}^{{{\log }_{2}}7}} \right)}^{2}}+{{\left( {{7}^{{{\log }_{7}}9}} \right)}^{2}}+{{\left( {{9}^{{{\log }_{9}}11}} \right)}^{2}}={{7}^{2}}+{{9}^{2}}+{{11}^{2}}=251.$
Đáp án C.