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Vì \(\frac{x}{3}\) = \(\frac{y}{4}\) => \(\frac{x}{15}\) = \(\frac{y}{20}\)
\(\frac{y}{5}\) = \(\frac{z}{6}\) => \(\frac{y}{20}\) = \(\frac{z}{24}\)
nên \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{24}\)
Đặt \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{24}\) = k
=> x = 15k; y = 20k và z = 24k
Thay vào M ta đc:
M = \(\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}\)
= \(\frac{30k+60k+96k}{45k+80k+120k}\)
= \(\frac{\left(30+60+96\right)k}{\left(45+80+120\right)k}\)
= \(\frac{186k}{245k}\) = \(\frac{186}{245}\)
Vậy M = \(\frac{186}{245}\).
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Vì \(\frac{3x-y}{x+3}=\frac{3}{4}=\frac{3x-y+3}{x+3+4}=\frac{3x+3-Y}{x+7}\)
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\(\frac{3x}{4}=\frac{4y}{5}=\frac{2z}{3}\)
vì \(4\ne5\ne3\)
nên \(\frac{3x}{4}=\frac{4y}{5}=\frac{2z}{3}=0\)
do đó 3x=4y=2z=0 nên x=y=z=0
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Theo đề, ta có:
\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{1}{3}x=-2t\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{x}{-2}=\dfrac{t}{\dfrac{1}{3}}\end{matrix}\right.\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{-\dfrac{1}{3}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{-\dfrac{1}{3}}=\dfrac{x+y+z-2t}{2+3+4-2\cdot\dfrac{-1}{3}}=\dfrac{4}{\dfrac{29}{3}}=\dfrac{12}{29}\)
Do đó: x=24/29; y=36/29; z=48/29; t=-4/29
\(\dfrac{x}{2}+\dfrac{y}{3}-z+t=\dfrac{12}{29}+\dfrac{12}{29}-\dfrac{48}{29}+\dfrac{-4}{29}=-\dfrac{28}{29}\)
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\(\frac{3x-y}{x+y}=\frac{3}{4}\)
<=> (3x - y) . 4 = (x + y) . 3
<=> 12x - 4y = 3x + 3y
<=> 12x - 3x = 3y + 4y
<=> 9x = 7y
<=> \(\frac{x}{y}=\frac{7}{9}\)
\(\frac{3x-y}{x+y}=\frac{3}{4}\)
=>(3x-y).4=(x+y).3
=>12x-4y=3x+3y
=>12x-4x=3y+4y
=>8x=7y
=>\(\frac{x}{y}=\frac{7}{8}\)
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C, CHO 7X=3Y VA X -Y =16
=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\)
=> \(\hept{\begin{cases}x=-4.3\\y=-4.7\end{cases}\Rightarrow\hept{\begin{cases}x=-12\\y=-28\end{cases}}}\)
bạn viết lại đề đi đè gì mà sai hết
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\(\frac{3x-y}{x+y}=\frac{3}{4}\\ \Rightarrow3\left(x+y\right)=4\left(3x-y\right)\\ \Rightarrow3x+3y=12x-4y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\frac{x}{y}=\frac{7}{9}\)
\(\frac{3x-y}{x+y}=\frac{3}{4}\)
\(\Rightarrow\left(3x-y\right).4=\left(x+y\right).3\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow12x-3x=3y+4y\)
\(\Rightarrow9x=7y\)
\(\Rightarrow\frac{x}{y}=\frac{7}{9}\)
Vậy \(\frac{x}{y}=\frac{7}{9}.\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{3x-y}{x+y}=\frac{3}{4}\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow12x-3x=3y+4y\)
\(\Rightarrow9x=7y\)
\(\Rightarrow\frac{x}{y}=\frac{7}{9}\)
\(\frac{3x-y}{x+y}=\frac{3}{4}\)
\(\Leftrightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(\Leftrightarrow12x-4y=3x+3y\)
\(\Leftrightarrow9x=7y\)
\(\Leftrightarrow\frac{x}{y}=\frac{7}{9}\)
\(\frac{3x-y}{x+y}=\frac{3}{4}\Leftrightarrow4(3x-y)=3(x+y)\)
\(\Leftrightarrow12x-4y=3x+3y\)
\(\Leftrightarrow12x-4y-3x=3y\)
\(\Leftrightarrow9x=7y\Leftrightarrow\frac{x}{y}=\frac{7}{9}\)