\(\frac{x}{y}=\frac{3}{5}\Leftrightarrow5x=3y\Leftrightarrow35x=21y\)

\(7y=6z\Leftrightarrow21y=18z\)

Suy ra \(35x=18z\)

\(4x+8y-9z=-3\)

\(40x+80y-90z=-30\)

\(5x+35x+80y-90z=-30\)

\(83y-72z=-30\)

\(83y-84y=-30\left(Vì6z=7y\Leftrightarrow-72z=-84y\right)\)

\(y=30\)

\(x=18\)

\(z=35\)

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

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

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

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

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

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:

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

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

bài 2 :

ta có x:y:z=3:5:(-2)

=>x/3=y/5=z/-2

=>5x/15=y/5=3z/-6

áp dụng tc dãy ... ta có :

5x/15=y/5=3z/-6=5x-y+3z/15-5+(-6)=-16/4=-4

=>x/3=-=>x=-12

=>y/5=-4=>y=-20

=>z/-2=-4=>z=8