a)\(\frac{x-2}{x-3}=\frac{x+3}{x+5}\Rightarrow\left(x-2\right)\left(x+5\right)=\left(x-3\right)\left(x+3\right)\)
\(\Rightarrow x^2+3x-10=x^2-9\)
\(\Rightarrow x^2+3x-10-x^2+9=0\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\frac{1}{3}\)
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b)Theo bài ra ta có:
\(xy=96;2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\)
Đặt \(\frac{x}{3}=\frac{y}{2}=k\Rightarrow x=3k;y=2k\)
\(\Rightarrow xy=96\Leftrightarrow3k\cdot2k=96\)
\(\Leftrightarrow6k^2=96\)
\(\Leftrightarrow k^2=16\Leftrightarrow k=\pm4\)
Nếu k=4 thì \(\hept{\begin{cases}x=3k=3\cdot4=12\\y=2k=2\cdot4=8\end{cases}}\)
Nếu k=-4 thì \(\hept{\begin{cases}x=3k=3\cdot\left(-4\right)=-12\\y=2k=2\cdot\left(-4\right)=-8\end{cases}}\)
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c)Theo bài ra ta có:
\(x-2y+z=34;5x=8y=3z\)\(\Leftrightarrow\frac{5x}{120}=\frac{8y}{120}=\frac{3z}{120}\Leftrightarrow\frac{x}{24}=\frac{y}{15}=\frac{z}{40}\)
\(\Leftrightarrow\frac{x}{24}=\frac{2y}{30}=\frac{z}{40}\)
Áp dụng tc dãy tỉ :
\(\frac{x}{24}=\frac{2y}{30}=\frac{z}{40}=\frac{x-2y+z}{24-30+40}=\frac{34}{34}=1\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{24}=1\Rightarrow24\\\frac{2y}{30}=1\Rightarrow y=\frac{30}{2}=15\\\frac{z}{40}=1\Rightarrow z=40\end{cases}}\)
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d)Theo bài ra ta có:
\(3x+5y+7z=123\);\(\frac{x}{2}=\frac{2y}{5}=\frac{4z}{7}\Leftrightarrow\frac{x}{8}=\frac{y}{10}=\frac{z}{7}\)
\(\Leftrightarrow\frac{3x}{24}=\frac{5y}{50}=\frac{7z}{49}\)
Áp dụng tc dãy tỉ:
\(\frac{3x}{24}=\frac{5y}{50}=\frac{7z}{49}=\frac{3x+5y+7z}{24+50+49}=\frac{123}{123}=1\)
\(\Rightarrow\hept{\begin{cases}\frac{3x}{24}=1\Rightarrow x=\frac{24}{3}=8\\\frac{5y}{50}=1\Rightarrow y=\frac{50}{5}=10\\\frac{7z}{49}=1\Rightarrow z=\frac{49}{7}=7\end{cases}}\)
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