Ta có:

\(\dfrac{x}{y}=\dfrac{3}{5}\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}\)

\(7y=6z\Rightarrow\dfrac{y}{30}=\dfrac{z}{35}\)

\(\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}\)

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

\(\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}=\dfrac{4x}{72}=\dfrac{8y}{240}=\dfrac{9z}{315}=\dfrac{4x+8y-9z}{72+240-315}=\dfrac{-3}{-3}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=18\\y=30\\z=35\end{matrix}\right.\)

Vậy...

Ta có: \(\dfrac{x}{y}=\dfrac{3}{5}\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\) (1)

\(7y=6z\Rightarrow\dfrac{y}{6}=\dfrac{z}{7}\) (2)

Từ (1) và (2) suy ra: \(\dfrac{x}{3}=\dfrac{y}{5};\dfrac{y}{6}=\dfrac{z}{7}\Leftrightarrow\dfrac{x}{18}=\dfrac{y}{30};\dfrac{y}{30}=\dfrac{z}{35}\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}\)

Có \(\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}\)và \(4x+8y-9z=-3\)

Áp dụng tính chất dãu tỉ số bằng nhau ta có:

\(\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}\Rightarrow\dfrac{4x}{72}=\dfrac{8y}{240}=\dfrac{9z}{315}=\dfrac{4x+8y-9z}{72+240-315}=\dfrac{-3}{-3}=1\)

\(\dfrac{4x}{72}=1\Rightarrow4x=72\Rightarrow x=\dfrac{72}{4}=18\)

\(\dfrac{8y}{240}=1\Rightarrow8y=240\Rightarrow y=\dfrac{240}{8}=30\)

\(\dfrac{9z}{315}=1\Rightarrow9z=315\Rightarrow z=\dfrac{315}{9}=35\)

Vậy x=18 ; y=30 ; z=35

Ta có:

\(\dfrac{x}{y}=\dfrac{3}{5}\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}\)

\(7y=6z\Rightarrow\dfrac{y}{30}=\dfrac{z}{35}\)

\(\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}\)

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

\(\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}=\dfrac{4x}{72}=\dfrac{8y}{240}=\dfrac{9z}{315}=\dfrac{4x+8y-9z}{72+240- 315}=\dfrac{-3}{-3}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=18\\y=30\\z=35\end{matrix}\right.\)

Vậy...

Ta có: \(\frac{x}{3}=\frac{y}{5};\frac{y}{6}=\frac{z}{7}\Rightarrow\frac{x}{18}=\frac{y}{30}=\frac{z}{35}=\frac{4x}{72}=\frac{8y}{240}=\frac{9z}{315}=\frac{-3}{-3}=1\)

\(\Rightarrow\frac{x}{18}=1\Rightarrow x=18;\frac{y}{30}=1\Rightarrow y=30;\frac{z}{35}=1\Rightarrow z=35\) 

a) \(\left(3x+y-z\right)-\left(4x-2y+6z\right)\)

\(=3x+y-z-4x+2y-6z\)

\(=-x+3y-7z\)

b) \(\left(x^3+6x^2+5y^3\right)-\left(2x^3-5x+7y^3\right)\)

\(=x^3+6x^2+5y^3-2x^3+5x-7y^3\)

\(=-x^3+6x^2+5x-2y^3\)

c) \(\left(5,7x^{2y}-3,1xy+8y^3\right)-\left(6,9xy-2,3x^{2y}-8y^3\right)\)

\(=5,7x^{2y}-3,1xy+8y^3-6,9xy+2,3x^{2y}+8y^3\)

\(=8x^{2y}-10xy+16y^3\)

Vì \(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\)

\(4x=6z\Rightarrow\frac{x}{6}=\frac{z}{4}\Rightarrow\frac{x}{3}=\frac{z}{2}\)

\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{2}\)

\(\Rightarrow\frac{2x}{6}=\frac{7y}{28}=\frac{3z}{6}\)

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

\(\frac{2x}{6}=\frac{7y}{28}=\frac{3z}{6}=\frac{2x+7y-3z}{6+28-6}=\frac{2}{28}=\frac{1}{14}\)

\(\cdot\frac{x}{3}=\frac{1}{14}\Rightarrow x=\frac{3}{14}\)

\(\cdot\frac{y}{4}=\frac{1}{14}\Rightarrow y=\frac{2}{7}\)

\(\cdot\frac{z}{2}=\frac{1}{14}\Rightarrow z=\frac{1}{7}\)

a) Theo bài ra , ta có : x : y : z = 3 : 5 : ( -2 )

=> \(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\) => \(\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}\) và 5x - y + 3z = -16

Áp dụng t/c của dãy tỉ số = nhau , ta có :

\(\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}=\frac{5x-y+3z}{15-5+\left(-6\right)}=\frac{-16}{-4}=4\)

\(\frac{x}{3}=4\Rightarrow x=4.3=12\\ \frac{y}{5}=4\Rightarrow y=4.5=20\\ \frac{z}{-2}=4\Rightarrow z=-2.4=-8\)

Vậy x = 12 ; y = 20 ; z = -8

a) Ta có : x : y : z = 3 : 5 : (-2) \(\Rightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\Rightarrow\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}\)

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

\(\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}=\frac{5x-y+3z}{15-5+-6}=-\frac{16}{4}=-4\)

\(\Rightarrow\begin{cases}\frac{5x}{15}=4\\\frac{y}{5}=4\\\frac{3z}{-6}=4\end{cases}\Rightarrow\begin{cases}5x=4.15\\y=4.5\\3z=4.\left(-6\right)\end{cases}\Rightarrow\begin{cases}5x=60\\y=20\\3z=-24\end{cases}\Rightarrow\begin{cases}x=12\\y=20\\z=-8\end{cases}\)

b) 2x = 3y \(\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\) (1)

5y = 7z \(\Rightarrow\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\Rightarrow\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}\)

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

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

\(\Rightarrow\begin{cases}\frac{3x}{63}=2\\\frac{7y}{98}=2\\\frac{5z}{50}=2\end{cases}\Rightarrow\begin{cases}3x=2.63\\7y=2.98\\5z=2.50\end{cases}\Rightarrow\begin{cases}3x=126\\7y=196\\5z=100\end{cases}\Rightarrow\begin{cases}x=42\\y=28\\z=20\end{cases}\)

c) x : y : z = 4 : 5 : 6 \(\Rightarrow\frac{x}{4}=\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{x^2}{16}=\frac{y^2}{25}=\frac{z^2}{36}\Rightarrow\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}\)

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

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

\(\Rightarrow\begin{cases}x^2=9.16\\2y^2=9.50\\z^2=9.36\end{cases}\Rightarrow\begin{cases}x^2=144\\y^2=450\div2=225\\z^2=324\end{cases}\Rightarrow\begin{cases}x=\pm12\\y=\pm15\\z=\pm18\end{cases}\)

Vậy x = 12 ; y = 15 ; z = 18

hoặc x = -12 ; y = -15 ; z = -18

b) Theo bài ra , ta có :

2x = 3y => \(\frac{x}{3}=\frac{y}{2}=\frac{x}{21}=\frac{y}{14}\) (1)

5y = 7z => \(\frac{y}{7}=\frac{z}{5}=\frac{y}{14}=\frac{z}{10}\) (2)

Tứ (1) , (2) => \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\) => \(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}\) và 3x - 7y + 5z = 30

Áp dụng t/c của dãy ti số = nhau , ta có :

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

\(\frac{x}{21}=2\Rightarrow x=2.21=42\\ \frac{y}{14}=2\Rightarrow y=2.14=28\\ \frac{z}{10}=2\Rightarrow z=2.10=20\\\)

Vậy x = 42 ; y = 28 ; z = 20

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:

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

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