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Ta có \(\dfrac{2x-3}{5}=\dfrac{3y+2}{7}=\dfrac{z-1}{3}=\dfrac{4x-6}{10}=\dfrac{6y+4}{14}=\dfrac{7z-7}{21}\)
Áp dụng t/c dtsbn:
\(\dfrac{4x-6}{10}=\dfrac{6y+4}{14}=\dfrac{7z-7}{21}=\dfrac{\left(4x-6y+7z\right)-6-4-7}{10-14+21}=\dfrac{68-17}{17}=3\\ \Rightarrow\left\{{}\begin{matrix}2x-3=15\\3y+2=21\\z-1=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\y=\dfrac{19}{3}\\z=10\end{matrix}\right.\)
a: Ta có: 2x/3=3y/4=4z/5
nên \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
Đặt \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=k\)
=>x=3/2k; y=4/3k; z=5/4k
\(xy+yz-xz=32\)
\(\Leftrightarrow\dfrac{3}{2}k\cdot\dfrac{4}{3}k+\dfrac{4}{3}k\cdot\dfrac{5}{4}k-\dfrac{3}{2}k\cdot\dfrac{5}{4}k=32\)
\(\Leftrightarrow k^2\cdot\dfrac{43}{24}=32\)
\(\Leftrightarrow k^2=\dfrac{768}{43}\)
Trường hợp 1: \(k=\dfrac{16\sqrt{129}}{43}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{24\sqrt{129}}{43}\\y=\dfrac{64\sqrt{129}}{129}\\z=\dfrac{20\sqrt{129}}{43}\end{matrix}\right.\)
Trường hợp 2: \(k=-\dfrac{16\sqrt{129}}{43}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{24\sqrt{129}}{43}\\y=-\dfrac{64\sqrt{129}}{129}\\z=-\dfrac{20\sqrt{129}}{43}\end{matrix}\right.\)
b: Ta có: 4x=3y
nên x/3=y/4=k
=>x=3k; y=4k
\(x^2-xy+y^2=32\)
\(\Leftrightarrow9k^2-12k^2+16k^2=32\)
\(\Leftrightarrow13k^2=32\)
Trường hợp 1: \(k=\dfrac{32\sqrt{13}}{13}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{96\sqrt{13}}{13}\\y=\dfrac{128\sqrt{13}}{13}\end{matrix}\right.\)
Trường hợp 2: \(k=-\dfrac{32\sqrt{13}}{13}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{96\sqrt{13}}{13}\\y=-\dfrac{128\sqrt{13}}{13}\end{matrix}\right.\)
a) \(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\Rightarrow\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)
Khi đó: \(\hept{\begin{cases}\frac{5x}{50}=2\Rightarrow x=20\\\frac{y}{6}=2\Rightarrow y=12\\\frac{2z}{42}=2\Rightarrow z=42\end{cases}}\)
e) \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z-5}{9}=\frac{50-5}{9}=5\)
Khi đó: \(\hept{\begin{cases}\frac{2x-2}{4}=5\Rightarrow x=11\\\frac{3y-6}{9}=5\Rightarrow y=17\\\frac{z-3}{4}=5\Rightarrow z=23\end{cases}}\).