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a/ \(\frac{x}{y}=5\Rightarrow\frac{x}{5}=\frac{y}{1}=\frac{x+y}{5+1}=\frac{18}{6}=3\)
=> x = 15 , y = 3
b/ \(\frac{x}{17}=\frac{y}{2}\Rightarrow\frac{2x}{34}=\frac{y}{2}=\frac{2x-y}{34-2}=\frac{64}{32}=2\)
=> x = 34, y = 4
c/ \(3x=7y\Rightarrow\frac{x}{7}=\frac{y}{3}=\frac{x-y}{7-3}=\frac{-16}{4}=-4\)
=> x = -28 , y=-12
d,e,f,g,h tương tự.
i/ \(x:y=5:6\Rightarrow\frac{x}{5}=\frac{y}{6}\)
Làm tương tự các câu còn lại.
j/ Đặt \(\frac{x}{4}=\frac{y}{7}=k\) \(\Rightarrow\begin{cases}x=4k\\y=7k\end{cases}\)
xy = 112 => 4k.7k = 112 => \(k^2=4\Rightarrow k=\pm2\)
Nếu k = 2 thì x = 8, y = 14
Nếu k = -2 thì x = -8 , y = -14
k/ \(-2x=3y\Rightarrow\frac{x}{3}=\frac{y}{-2}\)
Làm tương tự câu j.
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Áp dụng tc của dãy tỉ số bằng nhau ta có :
\(\frac{x}{20}=\frac{y}{15}=\frac{y-x}{15-20}=\frac{20}{-5}=-4\)
\(\Rightarrow\begin{cases}x=-80\\y=-60\end{cases}\)
Theo t/c của dãy tỉ số bằng nhau ta có
\(\frac{x}{20}=\frac{y}{15}=\frac{y-x}{15-20}=\frac{20}{-5}=-4\)
\(\frac{x}{20}=-4\Rightarrow x=-4.20=-80\)
\(\frac{y}{15}=-4\Rightarrow y=-4.15=-60\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: 2xx=3y=>x/3=y/2=>x/21=y/14 ; x/7=z/5=>x/21=z/15 =>x/21=y/14=z/15=>3x/63=7y/98=5z/75 ADTCDTSBN ta có 3x/63=7y/98=5z /75=3x-7y+5z=40/63-98+75=40=1 3x=1.63=63 =>x=21 ;7y=1.98=98=>y=14 ; 5z=1.75=>z=15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\Leftrightarrow\int^{\frac{2x}{30}=\frac{3y}{60}=\frac{4z}{96}}_{\frac{3x}{45}=\frac{4y}{80}=\frac{5z}{120}}\)
Theo t/c tỉ lệ tức ta có:
\(\frac{2x+3y+4z}{30+60+96}=\frac{3x+4y+5z}{45+80+120}\Leftrightarrow\frac{2x+3y+4z}{3x+4y+5z}=\frac{30+60+96}{45+80+120}=\frac{186}{245}\)
Vậy M=186/245
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
\(\begin{array}{l}\frac{x}{3} = \frac{y}{4} \Rightarrow \frac{x}{3}.\frac{1}{5} = \frac{y}{4}.\frac{1}{5} \Rightarrow \frac{x}{{15}} = \frac{y}{{20}};\\\frac{y}{5} = \frac{z}{6} \Rightarrow \frac{y}{5}.\frac{1}{4} = \frac{z}{6}.\frac{1}{4} \Rightarrow \frac{y}{{20}} = \frac{z}{{24}}\end{array}\)
Vậy \(\frac{x}{{15}} = \frac{y}{{20}} = \frac{z}{{24}}\) (đpcm)
b) Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{{15}} = \frac{y}{{20}} = \frac{z}{{24}} = \frac{{x - y + z}}{{15 - 20 + 24}} = \frac{{ - 76}}{{19}} = - 4\)
Vậy x = 15 . (-4) = -60; y = 20. (-4) = -80; z = 24 . (-4) = -96
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{3x}{4}\)= \(\frac{y}{2}\)= \(\frac{3z}{5}\)
=> \(\frac{1}{3}.\frac{3x}{4}=\frac{1}{3}.\frac{y}{2}=\frac{1}{3}.\frac{3z}{5}\)
\(\Rightarrow\frac{3x}{12}=\frac{y}{6}=\frac{3z}{15}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{4}=\frac{y}{6}=\frac{z}{5}=\frac{y-z}{6-5}=15\)
Suy ra:
- x = 15.4=60
- y=15.6=90
- z=15.5=75
\(\Rightarrow\)x + y + z = 225
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có 5-z=1 nên Z=4
Z=4 nên 4+Y=2 và Y=-2
Y=-2 nên -2+x=3 và X=5
Vậy...
báo cáo ko cút à
Cô giải cmnr
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x/y = -15/20 => x/-15 = y/20 => 3x/-45 = y/20
Áp dụng t/c của dãy tỉ số bằng nhau:
(tự túc là hạnh phúc ^^! Tự làm tiếp nhé!)
Ta đặt x=-15k
y=20k
nên 3x-y=-45k-20k=-130
-65k =-130
nên k=2
suy ra x=-30
y=40