6x=4y=z và 2x-3y+z=42 zvà tại sao lại được như vậy?
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a) 6x = 4y = z
\(\Rightarrow\frac{6x}{12}=\frac{4y}{12}=\frac{z}{12}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{12}=\frac{2x-3y+z}{4-9+12}=\frac{42}{7}=6\)
\(\Rightarrow\hept{\begin{cases}x=2.6=12\\y=3.6=18\\z=12.6=72\end{cases}}\)
\(\dfrac{x}{-3}=\dfrac{y}{5}\)⇒\(\dfrac{x}{-6}=\dfrac{y}{10}\)
\(\dfrac{y}{2}=\dfrac{z}{7}\)⇒\(\dfrac{y}{10}=\dfrac{z}{35}\)
⇒\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)
⇒\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\)
⇒\(\left\{{}\begin{matrix}x=-6.-6=36\\y=-6.10=-60\\z=-6.35=-210\end{matrix}\right.\)
\(a,\dfrac{x}{-3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{-6}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{7}\Rightarrow\dfrac{y}{10}=\dfrac{z}{35}\\ \Rightarrow\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}=\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\\ \Rightarrow\left\{{}\begin{matrix}x=36\\y=-60\\z=-210\end{matrix}\right.\)
\(b,6x=4y=z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y+z}{4-9+12}=\dfrac{42}{7}=6\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=18\\z=72\end{matrix}\right.\)
\(c,x=-2y\Rightarrow\dfrac{x}{-2}=y\Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}\\ 7y=2z\Rightarrow\dfrac{y}{2}=\dfrac{z}{7}\\ \Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}=\dfrac{2x}{-8}=\dfrac{3y}{6}=\dfrac{2x-3y+z}{-8+6+7}=\dfrac{42}{5}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{168}{5}\\y=\dfrac{84}{5}\\z=\dfrac{294}{5}\end{matrix}\right.\)
Bài 3 :
\(x=3y=2z\)
\(\Rightarrow x=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{2}}\)
\(\Rightarrow\frac{2x}{2}=\frac{3y}{1}=\frac{4z}{2}=\frac{2x-3y+4z}{2-1+2}=\frac{k}{3}\)
\(\Rightarrow x=\frac{k}{3}\)
\(y=\frac{k}{3}.\frac{1}{3}=\frac{k}{9}\)
\(z=\frac{k}{3}.\frac{1}{2}=\frac{k}{6}\)
Ta có : \(6x=4y\Rightarrow\frac{x}{4}=\frac{y}{6}\)(*)
\(4y=3z\Rightarrow\frac{y}{3}=\frac{z}{4}\)(**)
(*) => \(\frac{x}{12}=\frac{y}{18}\)(***)
(**) => \(\frac{y}{18}=\frac{z}{24}\)(****)
Từ (***) ; (****) suy ra : \(\frac{x}{12}=\frac{y}{18}=\frac{z}{24}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{12}=\frac{y}{18}=\frac{z}{24}=\frac{2x+3y-5z}{2.12+3.18-5.24}=-\frac{21}{-42}=\frac{1}{2}\)
\(\Rightarrow x=6;y=9;z=12\)
\(6x=4y=3z=t\Leftrightarrow x=\frac{t}{6},y=\frac{t}{4},z=\frac{t}{3}\).
\(2x+3y-5z=\frac{t}{3}+\frac{3t}{4}-\frac{5t}{3}=-\frac{7}{12}t=-21\Leftrightarrow t=36\)
\(\Rightarrow\hept{\begin{cases}x=\frac{36}{6}=6\\y=\frac{36}{4}=9\\z=\frac{36}{3}=12\end{cases}}\).