Giải bài 79 trang 127 SGK giải tích nâng cao 12
Giải các hệ phương trình:
a) \(\left\{ \begin{aligned} & {{3.2}^{x}}+{{2.3}^{y}}=2,75 \\ & {{2}^{x}}-{{3}^{y}}=-0,75 \\ \end{aligned} \right. \); b) \( \left\{ \begin{aligned} & {{\log }_{5}}x+{{\log }_{5}}7.{{\log }_{7}}y=1+{{\log }_{5}}2 \\ & 3+{{\log }_{2}}y={{\log }_{2}}5\left( 1+3{{\log }_{5}}x \right) \\ \end{aligned} \right. \)
a) \(\left\{ \begin{aligned} & {{3.2}^{x}}+{{2.3}^{y}}=2,75 \\ & {{2}^{x}}-{{3}^{y}}=-0,75 \\ \end{aligned} \right. \)
Đặt \(\left\{ \begin{aligned} & {{2}^{x}}=u \\ & {{3}^{y}}=v \\ \end{aligned} \right. \) hệ phương trình trở thành
\(\left\{ \begin{aligned} & 3u+2v=2,75 \\ & u-v=-0,75 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & u=\dfrac{1}{4} \\ & v=1 \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & {{2}^{x}}={{2}^{-2}} \\ & {{3}^{y}}=1={{3}^{0}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & x=-2 \\ & y=0 \\ \end{aligned} \right. \)
b) \( \left\{ \begin{aligned} & {{\log }_{5}}x+{{\log }_{5}}7.{{\log }_{7}}y=1+{{\log }_{5}}2 \\ & 3+{{\log }_{2}}y={{\log }_{2}}5\left( 1+3{{\log }_{5}}x \right) \\ \end{aligned} \right. \)
Điều kiện: \(x,y > 0\)
\(\Leftrightarrow \left\{ \begin{aligned} & {{\log }_{5}}x+{{\log }_{5}}y={{\log }_{5}}10 \\ & {{\log }_{2}}8+{{\log }_{2}}y={{\log }_{2}}5+3{{\log }_{2}}x \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & {{\log }_{5}}xy={{\log }_{5}}10 \\ &{{\log }_{2}}8y={{\log }_{2}}5{{x}^{3}} \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & xy=10 \\ & 8y=5{{x}^{3}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & y=\dfrac{10}{x} \\ & \dfrac{80}{x}=5{{x}^{3}} \\ \end{aligned} \right. \\ \Leftrightarrow \left\{ \begin{aligned} & y=\dfrac{10}{x} \\ & {{x}^{4}}=16 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & x=\pm 2 \\ & y=\pm 5 \\ \end{aligned} \right. \)
Do \(x,y>0\Rightarrow \left\{ \begin{aligned} & x=2 \\ & y=5 \\ \end{aligned} \right. \)
Ghi nhớ: Giả sử a, b, c > 0 và \(a\ne 1\). Ta có
\(\log_{a}\left( bc \right)={\log }_{a}{b}+{\log }_{a}{c}; \,\,\,\,\,\, {\log _{a}\left( \dfrac{b}{c} \right)={\log }_{a}}{{b}}-{\log }_{a}{c} \)
\({{\log }_{a}}\left( {{b}^{\alpha }} \right)=\alpha {{\log }_{a}}b;\,\,\,\,\,\,\,{{a}^{{{\log }_{a}}b}}=b;\,\,\,\,\,\)