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Ta có: \(\frac{7x+5y}{3x-7y}=\frac{7z+5t}{3z-7t}\)
\(\Rightarrow\) \(\frac{7x+5y}{7z+5t}=\frac{3x-7y}{3z-7t}\)
+) \(\left[{}\begin{matrix}\frac{7x+5y}{7z+5t}=\frac{7x}{7z}=\frac{5y}{5t}=\frac{x}{z}=\frac{y}{t}\Rightarrow\frac{x}{y}=\frac{z}{t}\\\frac{3x-7y}{3z-7t}=\frac{3x}{3z}=\frac{7y}{7t}=\frac{x}{z}=\frac{y}{t}\Rightarrow\frac{x}{y}=\frac{z}{t}\end{matrix}\right.\)
\(\Rightarrow\frac{x}{y}=\frac{z}{t}\)
Vậy từ \(\frac{7x+5y}{3x-7y}=\frac{7z+5t}{3z-7t}\)\(\Rightarrow\frac{x}{y}=\frac{z}{t}\)
Tớ không biết trình bày có đúng không. Chúc bạn học tốt
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\(\frac{7x+5y}{3x-7y}=\frac{7z+5t}{3z-7t}\Rightarrow\left(7x+5y\right)\left(3z-7t\right)=\left(7z+5t\right)\left(3x-7y\right)\)
\(\Leftrightarrow21xz-49xt+15yz-35yt=21xz-49yz+15xt-35yt\)
\(\Leftrightarrow-49xt+15yz=-49yz+15xt\Leftrightarrow-64xt=-64yz\Leftrightarrow xt=yz\Leftrightarrow\frac{x}{y}=\frac{z}{t}\)
Vậy ta có đpcm
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\(\frac{7x+5y}{3x-7y}=\frac{7z+5t}{3z-7t}\Rightarrow\left(7x+5y\right)\left(3z-7t\right)=\left(7z+5t\right)\left(3x-7y\right)\)
\(\Leftrightarrow21xz+15yz-49tx-35ty=21xz+15tx-49yz-35ty\)
\(\Leftrightarrow21xz-21xz+15yz+49yz-49tx-15tx-35ty+35ty=0\)
\(\Leftrightarrow64yz-64tx=0\)
\(\Leftrightarrow yz=tx\)
\(\Leftrightarrow\frac{x}{y}=\frac{z}{t}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\dfrac{3x-5y}{2}=\dfrac{7y-3z}{3}=\dfrac{5z-7x}{4}=\dfrac{21x-35y}{14}=\dfrac{35y-15z}{15}=\dfrac{15z-21x}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{3x-5y}{2}=\dfrac{7y-3z}{3}=\dfrac{5z-7x}{4}=\dfrac{21x-35y}{14}=\dfrac{35y-15z}{15}=\dfrac{15z-21x}{12}=\dfrac{21x-35y+35y-15z+15z-21x}{14+15+12}=\dfrac{0}{41}=0\)
=>3x-5y=7y-3z=5z-7x=0
3x-5y=0 <=> 3x=5y <=> \(\dfrac{x}{5}=\dfrac{y}{3}\) (1)
7y-3z=0 <=> 7y=3z <=> \(\dfrac{y}{3}=\dfrac{z}{7}\) (2)
Từ (1) và (2) suy ra \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{7}\)
Áp dụng tính chất của dãy tỉ số bằng nhau: \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{7}=\dfrac{x+y+z}{5+3+7}=\dfrac{17}{15}\)
=>\(x=\dfrac{17}{15}.5=\dfrac{17}{3};y=\dfrac{17}{15}.3=\dfrac{17}{5};z=\dfrac{17}{15}.7=\dfrac{119}{15}\)
Vậy ...........
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Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{3x-5y}{2}=\frac{7y-3z}{3}=\frac{5z-7x}{4}\)
\(\Leftrightarrow \frac{7(3x-5y)}{14}=\frac{5(7y-3z)}{15}=\frac{3(5z-7x)}{12}=\frac{7(3x-5y)+5(7y-3z)+3(5z-7x)}{14+15+12}=0\)
Suy ra:
\(\left\{\begin{matrix} 3x=5y\\ 7y=3z\\ 5z=7x\end{matrix}\right.\Leftrightarrow 21x=35y=15z\)
\(\Leftrightarrow \frac{x}{\frac{1}{21}}=\frac{y}{\frac{1}{35}}=\frac{z}{\frac{1}{15}}=\frac{x+y+z}{\frac{1}{21}+\frac{1}{35}+\frac{1}{15}}=119\) (ADTCDTSBN)
\(\Rightarrow \left\{\begin{matrix} x=\frac{17}{3}\\ y=\frac{17}{5}\\ z=\frac{119}{15}\end{matrix}\right.\)
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\(2x=3y\text{⇒}\dfrac{x}{3}=\dfrac{y}{2}\text{⇒}\dfrac{x}{21}=\dfrac{y}{14}\)
\(5y=7z\text{⇒}\dfrac{y}{7}=\dfrac{z}{5}\text{⇒}\dfrac{y}{14}=\dfrac{z}{10}\)
⇒\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)⇒\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)
⇒x=42,y=28,z=20
\(\dfrac{x}{3}=\dfrac{y}{2}\)⇒\(\dfrac{x}{15}=\dfrac{y}{10}\)
\(\dfrac{x}{5}=\dfrac{z}{7}\text{⇒}\dfrac{x}{15}=\dfrac{z}{21}\)
⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{21}\)⇒\(\dfrac{x}{15}=\dfrac{2y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{2y}{20}=\dfrac{x+2y}{15+20}=\dfrac{-112}{35}=\dfrac{-16}{5}\)
⇒x=48,y=32,z=336/5
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4: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{38}{-19}=-2\)
Do đó: x=-16; y=-24; z=-30
Lời giải:
$\frac{7x+5y}{3x-5y}=\frac{7z+5t}{3z-5t}$
$\Rightarrow (7x+5y)(3z-5t)=(7z+5t)(3x-5y)$
$\Rightarrow 21xz-35xt+15yz-25yt = 21xz-35yz+15xt-25yt$
$\Rightarrow -35xt+15yz=-35yz+15xt$
$\Rightarrow -50xt=-50yz$
$\Rightarrow xt=yz\Rightarrow \frac{x}{y}=\frac{z}{t}$