Tìm ba số biết: (3a-2b)/5=(2c-5a)/3=(5b-3c)/2 và a^2+275=bc Giúp em với ạ. Em đang cần gấp. 27/07/2021 Bởi Mary Tìm ba số biết: (3a-2b)/5=(2c-5a)/3=(5b-3c)/2 và a^2+275=bc Giúp em với ạ. Em đang cần gấp.
Giải thích các bước giải: $\begin{array}{l}\frac{{3a – 2b}}{5} = \frac{{2c – 5a}}{3} = \frac{{5b – 3c}}{2}\\ \Rightarrow \left\{ \begin{array}{l}\frac{{34a}}{{15}} = \frac{{2c}}{3} + \frac{{2b}}{5}\\\frac{{13c}}{6} = \frac{{5b}}{2} + \frac{{5a}}{3}\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}a = \frac{{5c}}{{17}} + \frac{{3b}}{{17}}\\\frac{{13c}}{6} = \frac{{5b}}{2} + \frac{{5a}}{3}\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}a = \frac{{5c}}{{17}} + \frac{{3b}}{{17}}\\\frac{{13c}}{6} = \frac{{5b}}{2} + \frac{5}{3}(\frac{{5c}}{{17}} + \frac{{3b}}{{17}})\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}a = \frac{{5c}}{{17}} + \frac{{3b}}{{17}}\\\frac{{57c}}{{34}} = \frac{{95b}}{{34}}\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}a = \frac{{5c}}{{17}} + \frac{{3b}}{{17}}\\c = \frac{{5b}}{3}\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}a = \frac{5}{{17}}.\frac{{5b}}{3} + \frac{{3b}}{{17}}\\c = \frac{{5b}}{3}\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}a = \frac{{2b}}{3}\\c = \frac{{5b}}{3}\end{array} \right.\end{array}$ Vì ${a^2} + 275 = bc$ nên: $\begin{array}{l}{\frac{{2b}}{3}^2} + 275 = b.\frac{{5b}}{3}\\ \Leftrightarrow \frac{{11b}}{9} = 275\\ \Leftrightarrow b = 225\\ \Rightarrow a = 150,c = 375\end{array}$ Bình luận
Giải thích các bước giải:
$\begin{array}{l}
\frac{{3a – 2b}}{5} = \frac{{2c – 5a}}{3} = \frac{{5b – 3c}}{2}\\
\Rightarrow \left\{ \begin{array}{l}
\frac{{34a}}{{15}} = \frac{{2c}}{3} + \frac{{2b}}{5}\\
\frac{{13c}}{6} = \frac{{5b}}{2} + \frac{{5a}}{3}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = \frac{{5c}}{{17}} + \frac{{3b}}{{17}}\\
\frac{{13c}}{6} = \frac{{5b}}{2} + \frac{{5a}}{3}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = \frac{{5c}}{{17}} + \frac{{3b}}{{17}}\\
\frac{{13c}}{6} = \frac{{5b}}{2} + \frac{5}{3}(\frac{{5c}}{{17}} + \frac{{3b}}{{17}})
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = \frac{{5c}}{{17}} + \frac{{3b}}{{17}}\\
\frac{{57c}}{{34}} = \frac{{95b}}{{34}}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = \frac{{5c}}{{17}} + \frac{{3b}}{{17}}\\
c = \frac{{5b}}{3}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = \frac{5}{{17}}.\frac{{5b}}{3} + \frac{{3b}}{{17}}\\
c = \frac{{5b}}{3}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = \frac{{2b}}{3}\\
c = \frac{{5b}}{3}
\end{array} \right.
\end{array}$
Vì ${a^2} + 275 = bc$ nên:
$\begin{array}{l}
{\frac{{2b}}{3}^2} + 275 = b.\frac{{5b}}{3}\\
\Leftrightarrow \frac{{11b}}{9} = 275\\
\Leftrightarrow b = 225\\
\Rightarrow a = 150,c = 375
\end{array}$