(5-2x)/3=(4x-1)/-5 (12-3x)/32=6/(4-x) (10-2x)/6=27/(5-x) 17/07/2021 Bởi Jade (5-2x)/3=(4x-1)/-5 (12-3x)/32=6/(4-x) (10-2x)/6=27/(5-x)
Đáp án: $\begin{array}{l}a)\dfrac{{5 – 2x}}{3} = \dfrac{{4x – 1}}{{ – 5}}\\ \Rightarrow – 5.\left( {5 – 2x} \right) = 3.\left( {4x – 1} \right)\\ \Rightarrow – 25 + 10x = 12x – 3\\ \Rightarrow 12x – 10x = – 25 + 3\\ \Rightarrow 2x = – 22\\ \Rightarrow x = – 11\\Vay\,x = – 11\\)\dfrac{{12 – 3x}}{{32}} = \dfrac{6}{{4 – x}}\\ \Rightarrow 3\left( {4 – x} \right)\left( {4 – x} \right) = 6.32\\ \Rightarrow {\left( {4 – x} \right)^2} = 64\\ \Rightarrow \left[ \begin{array}{l}4 – x = 8\\4 – x = – 8\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = – 4\\x = 12\end{array} \right.\\Vay\,x = – 4\,hoac\,x = 12\\c)\dfrac{{10 – 2x}}{6} = \dfrac{{27}}{{5 – x}}\\ \Rightarrow \left( {10 – 2x} \right)\left( {5 – x} \right) = 6.27\\ \Rightarrow 2{\left( {5 – x} \right)^2} = 6.27\\ \Rightarrow {\left( {5 – x} \right)^2} = 81\\ \Rightarrow \left[ \begin{array}{l}5 – x = 9\\5 – x = – 9\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = – 4\\x = 14\end{array} \right.\\Vay\,x = – 4\,hoac\,14\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)\dfrac{{5 – 2x}}{3} = \dfrac{{4x – 1}}{{ – 5}}\\
\Rightarrow – 5.\left( {5 – 2x} \right) = 3.\left( {4x – 1} \right)\\
\Rightarrow – 25 + 10x = 12x – 3\\
\Rightarrow 12x – 10x = – 25 + 3\\
\Rightarrow 2x = – 22\\
\Rightarrow x = – 11\\
Vay\,x = – 11\\
)\dfrac{{12 – 3x}}{{32}} = \dfrac{6}{{4 – x}}\\
\Rightarrow 3\left( {4 – x} \right)\left( {4 – x} \right) = 6.32\\
\Rightarrow {\left( {4 – x} \right)^2} = 64\\
\Rightarrow \left[ \begin{array}{l}
4 – x = 8\\
4 – x = – 8
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = – 4\\
x = 12
\end{array} \right.\\
Vay\,x = – 4\,hoac\,x = 12\\
c)\dfrac{{10 – 2x}}{6} = \dfrac{{27}}{{5 – x}}\\
\Rightarrow \left( {10 – 2x} \right)\left( {5 – x} \right) = 6.27\\
\Rightarrow 2{\left( {5 – x} \right)^2} = 6.27\\
\Rightarrow {\left( {5 – x} \right)^2} = 81\\
\Rightarrow \left[ \begin{array}{l}
5 – x = 9\\
5 – x = – 9
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = – 4\\
x = 14
\end{array} \right.\\
Vay\,x = – 4\,hoac\,14
\end{array}$