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2x=3y;5y=7z
=>x/3=y/2;y/7=z/5
=>x/21=x/14;y/14=z/10
=>x/21=y/14=z/10
=>3x/63=7y/98=5z/50
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
3x/63=7y/98=5z/50=3x-7y+5z/63-98+50=30/15=2
suy ra : 3x/63=2 =>3x=126 =>x=126:3=42

:D
7y/98=2 =>7y =196 =>y=196:7=28
5z/50=2 =>5z = 100 => z=100:5=20

ai do biet giup minh voi

a)Ta có: \(2x=3y;5y=7z\)và \(x-y-z=-27\)

\(\Rightarrow\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\)và\(x-y-z=-27\)

\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)và \(x-y-z=-27\)

Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{x-y-z}{21-14-10}=\frac{-27}{-3}=9\)

Ta có:\(\frac{x}{21}=9\Rightarrow x=9.21=189\)

          \(\frac{y}{14}=9\Rightarrow y=9.14=126\)

         \(\frac{z}{10}=9\Rightarrow z=9.10=90\)

Vậy:\(x=189;y=126\)và\(z=90\)

b) \(\frac{x}{4}=\frac{y}{5}=\frac{z}{6}\)và\(x^2-2y^2+z^2=18\)

\(\Rightarrow\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}\)và\(x^2-2y^2+z^2=18\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}=\frac{x^2-2y^2+z^2}{16-50+36}=\frac{18}{2}=9\)

Ta có:\(\frac{x^2}{16}=9\Rightarrow x^2=144\Rightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

\(\frac{2y^2}{50}=9\Rightarrow2y^2=450\Rightarrow y^2=225\Rightarrow\orbr{\begin{cases}y=15\\y=-15\end{cases}}\)

\(\frac{z^2}{36}=9\Rightarrow z^2=324\Rightarrow\orbr{\begin{cases}z=18\\z=-18\end{cases}}\)

Vậy: \(x=12;y=15;z=18\)hoặc \(x=-12;y=-15;z=-18\)

c) \(x:y:z=3:8:5\)và\(3x+y-2z=14\)

\(\Rightarrow\frac{x}{3}=\frac{y}{8}=\frac{z}{5}\)và\(3x+y-2z=14\)

\(\Rightarrow\frac{3x}{9}=\frac{y}{8}=\frac{2z}{10}\)và \(3x+y-2z=14\)

Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:

\(\frac{3x}{9}=\frac{y}{8}=\frac{2z}{10}=\frac{3x+y-2z}{9+8-10}=\frac{14}{7}=2\)

Ta có: \(\frac{3x}{9}=2\Rightarrow3x=18\Rightarrow x=6\)

\(\frac{y}{8}=2\Rightarrow y=16\)

\(\frac{2z}{10}=2\Rightarrow2z=20\Rightarrow z=10\)

Vậy:\(x=6;y=16;z=10\)

Áp dụng t/c dtsbn:

\(\dfrac{x+3}{5}=\dfrac{y-2}{3}=\dfrac{z-1}{7}=\dfrac{3x+9}{15}=\dfrac{5y-10}{15}=\dfrac{7z-7}{49}=\dfrac{3x-5y+7z+9+10-7}{15-15+49}=\dfrac{86+12}{49}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x+3=2.5=10\\y-2=2.3=6\\z-1=2.7=14\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=10-3=7\\y=6+2=8\\z=14+1=15\end{matrix}\right.\)

e) Ta có:

\(\left\{{}\begin{matrix}2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{1}{7}.\frac{x}{3}=\frac{1}{7}.\frac{y}{2}\Leftrightarrow\frac{x}{21}=\frac{y}{14}\\7z=5y\Leftrightarrow\frac{z}{5}=\frac{y}{7}\Leftrightarrow\frac{1}{2}.\frac{z}{5}=\frac{1}{2}.\frac{y}{7}\Leftrightarrow\frac{z}{10}=\frac{y}{14}\end{matrix}\right.\)

\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=42\\y=28\\z=20\end{matrix}\right.\)

f)Ta có:

\(\frac{x}{4}=\frac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)

\(\Rightarrow xy=4k5k=20k^2=80\Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

TH1: \(k=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)

TH2: \(k=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-10\end{matrix}\right.\)

g)Ta có:

\(\frac{x+3}{5}=\frac{y-2}{3}=\frac{z-1}{7}=\frac{3\left(x+3\right)}{15}=\frac{5\left(y-2\right)}{15}=\frac{7\left(z-1\right)}{49}=\frac{3x+9}{15}=\frac{5y-10}{15}=\frac{7z-7}{49}=\frac{3x+9+5y-10-\left(7z-7\right)}{15+15-49}=\frac{3x+5y-7z+\left(9-10+7\right)}{-19}=\frac{38}{-19}=-2\)

đặt x+3/5=y-2/3=z-1/7=k

=> x=5k-3  ; y=3k+2   ;  z=7k+1

ta có:

3(5k-3)-5(3k+2)+7(7k+1)=86

15k-9-15k-10+49k+7=86

49k-12=86

49k=98

k=2

ta có: x=2x5-3=7

y=2x3+2=8

z=2x7+1=15

1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)

2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)

3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)

a) Áp dụng tính chất dãy tỉ số bằng nhau ta có : \(\frac{x}{20}=\frac{y}{9}=\frac{z}{6}=\frac{x-2y+4z}{20-2.9+4.6}=\frac{13}{26}=\frac{1}{2}\)

* \(\frac{x}{20}=\frac{1}{2}\Rightarrow x=\frac{1}{2}.20=10\)

*\(\frac{y}{9}=\frac{1}{2}\Rightarrow y=\frac{1}{2}.9=\frac{9}{2}\)

*\(\frac{z}{6}=\frac{1}{2}\Rightarrow z=\frac{1}{2}.6=3\)

b)c) đề bn viết ko rõ