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Câu hỏi: Tổng S tất cả các nghiệm của phương trình ${{7}^{3x}}-{{3.49}^{x}}{{.3}^{x}}+{{8.63}^{x}}-{{6.27}^{x}}=0$ là
A. $S=-1.$
B. $S=0.$
C. $S=1.$
D. $S=-4.$
A. $S=-1.$
B. $S=0.$
C. $S=1.$
D. $S=-4.$
Ta có: ${{7}^{3x}}-{{3.49}^{x}}{{.3}^{x}}+{{8.63}^{x}}-{{6.27}^{x}}=0\Leftrightarrow {{\left( {{7}^{x}} \right)}^{3}}-3.{{\left( {{7}^{x}} \right)}^{2}}{{.3}^{x}}+{{8.7}^{x}}.{{\left( {{3}^{x}} \right)}^{2}}-6.{{\left( {{3}^{x}} \right)}^{3}}=0$
$\Leftrightarrow \dfrac{{{\left( {{7}^{x}} \right)}^{3}}}{{{\left( {{3}^{x}} \right)}^{3}}}-3.\dfrac{{{\left( {{7}^{x}} \right)}^{2}}{{.3}^{x}}}{{{\left( {{3}^{x}} \right)}^{3}}}+8.\dfrac{{{7}^{x}}.{{\left( {{3}^{x}} \right)}^{2}}}{{{\left( {{3}^{x}} \right)}^{3}}}-6=0\Leftrightarrow {{\left( \dfrac{7}{3} \right)}^{3x}}-3.{{\left( \dfrac{7}{3} \right)}^{2x}}+8.{{\left( \dfrac{7}{3} \right)}^{x}}-6=0\Leftrightarrow {{\left( \dfrac{7}{3} \right)}^{x}}=1\Leftrightarrow x=0$
Vậy $S=0$
$\Leftrightarrow \dfrac{{{\left( {{7}^{x}} \right)}^{3}}}{{{\left( {{3}^{x}} \right)}^{3}}}-3.\dfrac{{{\left( {{7}^{x}} \right)}^{2}}{{.3}^{x}}}{{{\left( {{3}^{x}} \right)}^{3}}}+8.\dfrac{{{7}^{x}}.{{\left( {{3}^{x}} \right)}^{2}}}{{{\left( {{3}^{x}} \right)}^{3}}}-6=0\Leftrightarrow {{\left( \dfrac{7}{3} \right)}^{3x}}-3.{{\left( \dfrac{7}{3} \right)}^{2x}}+8.{{\left( \dfrac{7}{3} \right)}^{x}}-6=0\Leftrightarrow {{\left( \dfrac{7}{3} \right)}^{x}}=1\Leftrightarrow x=0$
Vậy $S=0$
Đáp án B.