$\frac{a}{3}$ + b = 15 b - c = $\frac{a}{4}$ $\frac{c}{4}$ + b = d a + b + c + d = 44 Xác định hệ số a, b, c, d. Giúp mình với ạ

Đáp án: $a = \dfrac{{366}}{{11}};b = \dfrac{{43}}{{11}};c =  - \dfrac{{97}}{{22}};d = \dfrac{{247}}{{22}}$

Giải thích các bước giải:

$\begin{array}{l}
 + )\dfrac{a}{3} + b = 15\\
 \Leftrightarrow \dfrac{a}{3} = 15 - b\\
 \Leftrightarrow a = 45 - 3b\\
 + b - c = \dfrac{a}{4}\\
 \Leftrightarrow c = b - \dfrac{a}{4} = b - \dfrac{{45 - 3b}}{4} = \dfrac{{7b - 45}}{4}\\
 + \dfrac{c}{4} + b = d\\
 \Leftrightarrow d = \dfrac{{\dfrac{{7b - 45}}{4}}}{4} + b = \dfrac{{23b - 45}}{4}\\
 \Leftrightarrow \left\{ \begin{array}{l}
a = 45 - 3b\\
c = \dfrac{{7b - 45}}{4}\\
d = \dfrac{{23b - 45}}{4}
\end{array} \right.\\
a + b + c + d = 44\\
 \Leftrightarrow 45 - 3b + b + \dfrac{{7b - 45}}{4} + \dfrac{{23b - 45}}{4} = 44\\
 \Leftrightarrow 45 - 2b + \dfrac{7}{4}b - \dfrac{{45}}{4} + \dfrac{{23}}{4}b - \dfrac{{45}}{4} = 44\\
 \Leftrightarrow \dfrac{{11}}{2}b = 44 - \dfrac{{45}}{2}\\
 \Leftrightarrow \dfrac{{11}}{2}b = \dfrac{{88 - 45}}{2} = \dfrac{{43}}{2}\\
 \Leftrightarrow b = \dfrac{{43}}{{11}}\\
 \Leftrightarrow \left\{ \begin{array}{l}
a = 45 - 3b = \dfrac{{366}}{{11}}\\
c = \dfrac{{7b - 45}}{4} = \dfrac{{ - 97}}{{22}}\\
d = \dfrac{{23b - 45}}{4} = \dfrac{{247}}{{22}}
\end{array} \right.\\
Vậy\,a = \dfrac{{366}}{{11}};b = \dfrac{{43}}{{11}};c =  - \dfrac{{97}}{{22}};d = \dfrac{{247}}{{22}}
\end{array}$