Đáp án: ${x = \dfrac{{149}}{{11}}y = \dfrac{{349}}{{33}};z = \dfrac{{170}}{{11}}}$ Giải thích các bước giải: Áp dụng tính chất dãy tỉ số bằng nhau ta có: $\begin{array}{l}\dfrac{{2x – 3}}{5} = \dfrac{{3y + 2}}{7} = \dfrac{{z – 1}}{3} = \dfrac{{2\left( {2x – 3} \right) – 2\left( {3y + 2} \right) + 5\left( {z – 1} \right)}}{{2.5 – 2.7 + 5.3}} = \dfrac{{4x – 6y + 5z – 15}}{{11}} = \dfrac{{68 – 15}}{{11}} = \dfrac{{53}}{{11}}\\ \Rightarrow \left\{ \begin{array}{l}\dfrac{{2x – 3}}{5} = \dfrac{{53}}{{11}}\\\dfrac{{3y + 2}}{7} = \dfrac{{53}}{{11}}\\\dfrac{{z – 1}}{3} = \dfrac{{53}}{{11}}\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}2x – 3 = \dfrac{{265}}{{11}}\\3y + 2 = \dfrac{{371}}{{11}}\\z – 1 = \dfrac{{159}}{{11}}\end{array} \right.\\ \Rightarrow \left\{ {x = \dfrac{{149}}{{11}}y = \dfrac{{349}}{{33}};z = \dfrac{{170}}{{11}}} \right.\end{array}$ Vậy ${x = \dfrac{{149}}{{11}}y = \dfrac{{349}}{{33}};z = \dfrac{{170}}{{11}}}$ Reply
Đáp án:
${x = \dfrac{{149}}{{11}}y = \dfrac{{349}}{{33}};z = \dfrac{{170}}{{11}}}$
Giải thích các bước giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\begin{array}{l}
\dfrac{{2x – 3}}{5} = \dfrac{{3y + 2}}{7} = \dfrac{{z – 1}}{3} = \dfrac{{2\left( {2x – 3} \right) – 2\left( {3y + 2} \right) + 5\left( {z – 1} \right)}}{{2.5 – 2.7 + 5.3}} = \dfrac{{4x – 6y + 5z – 15}}{{11}} = \dfrac{{68 – 15}}{{11}} = \dfrac{{53}}{{11}}\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{2x – 3}}{5} = \dfrac{{53}}{{11}}\\
\dfrac{{3y + 2}}{7} = \dfrac{{53}}{{11}}\\
\dfrac{{z – 1}}{3} = \dfrac{{53}}{{11}}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x – 3 = \dfrac{{265}}{{11}}\\
3y + 2 = \dfrac{{371}}{{11}}\\
z – 1 = \dfrac{{159}}{{11}}
\end{array} \right.\\
\Rightarrow \left\{ {x = \dfrac{{149}}{{11}}y = \dfrac{{349}}{{33}};z = \dfrac{{170}}{{11}}} \right.
\end{array}$
Vậy ${x = \dfrac{{149}}{{11}}y = \dfrac{{349}}{{33}};z = \dfrac{{170}}{{11}}}$