\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)

\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)

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

\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)

\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)

mọi người giúp mk câu b, c, d còn lại nha

\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)

\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)

\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)

\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)

a: \(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\left(3x-y\right)\cdot\left(3x+y\right)=9x^2-y^2\)

b: \(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\cdot\dfrac{1}{2}\)

\(=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\)

\(=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

c: \(\left(5y-3x\right)\cdot\dfrac{1}{4}\cdot\left(12x+20y\right)\)

\(=\left(5y-3x\right)\left(5y+3x\right)\)

\(=25y^2-9x^2\)

d: \(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(\dfrac{3}{2}y+x\right)\cdot2\)

\(=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)\)

\(=\dfrac{9}{4}y^2-x^2\)

e: \(\left(a+b+c\right)\left(a+b-c\right)\)

\(=\left(a+b\right)^2-c^2\)

\(=a^2+2ab+b^2-c^2\)

3x = 5y và 2y = -3z

\(\Rightarrow\frac{x}{6}=\frac{y}{10}=\frac{z}{-15}=\frac{x+y-z}{6+10-15}=\frac{2}{1}=2.\)

\(\frac{x}{6}=2\Rightarrow x=12\)

\(\frac{y}{10}=2\Rightarrow y=20\)

\(\frac{z}{15}=2\Rightarrow z=30\)

tham khảo nha 

Ta có: \(\hept{\begin{cases}3x=5y\\2y=-3z\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{5}=\frac{y}{3}\\\frac{y}{-3}=\frac{z}{2}\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{-15}=\frac{y}{-9}\\\frac{y}{-9}=\frac{z}{6}\end{cases}}}\)

\(\Rightarrow\frac{x}{-15}=\frac{y}{-9}=\frac{z}{6}\)

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

\(\frac{x}{-15}=\frac{y}{-9}=\frac{z}{6}=\frac{x+y-z}{-15-9-6}=\frac{2}{-30}=\frac{-1}{15}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{-1}{15}.\left(-15\right)=1\\y=\frac{-1}{15}.\left(-9\right)=\frac{3}{5}\\z=\frac{-1}{15}.6=\frac{-2}{5}\end{cases}}\)

Vậy ...

Từ 3x = 5y 

\(\Rightarrow\frac{x}{5}=\frac{y}{3}\)

Lại có : 2y = -3z

=)))))))))))))))))))))))))))))))))))))))))))))))))))))))

\(\dfrac{3x-2y}{5}\)=\(\dfrac{2z-5x}{3}\)=\(\dfrac{5y-3z}{2}\)

⇒\(\dfrac{15x-10y}{25}\)=\(\dfrac{6z-15x}{9}\)=\(\dfrac{10y-6z}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{15x-10y}{25}\)=\(\dfrac{6z-15x}{9}\)=\(\dfrac{10y-6z}{4}\)=\(\dfrac{15x-10y+6z-15x+10y-6z}{25+9+4}\)=0

⇒3x-2y=2z-5x=5y-3z=0

* 3x-2y=0⇒3x=2y⇒\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\) 

* 2z-5x=0⇒2z=5x⇒\(\dfrac{z}{5}\)=\(\dfrac{x}{2}\) 

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\)=\(\dfrac{x+y+z}{2+3+5}\)=\(\dfrac{-50}{10}\)=-5

\(\dfrac{x}{2}\)=-5⇒x=-10

\(\dfrac{y}{3}\)=-5⇒y=-15

\(\dfrac{z}{5}\)=-5⇒z=-25

Vậy x=-10;y=-15;z=-25

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

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

\(\hept{\begin{cases}\frac{x}{2}=\frac{x}{3}\\\frac{y}{5}=\frac{x}{7}\end{cases}\Rightarrow}\frac{x}{2}=\frac{5y}{15};\frac{3y}{15}=\frac{z}{7}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

Áp dụng tính chát dãy tỉ số = nhau ta có:

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)

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

\(\frac{y}{15}=2\Rightarrow y=30\)

\(\frac{z}{21}=3\Rightarrow z=63\)

b, Tự làm

c, \(5x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{5}\)

\(2x=3z\Leftrightarrow\frac{x}{3}=\frac{z}{2}\)

\(\Leftrightarrow\frac{x}{2}=\frac{y}{5};\frac{x}{3}=\frac{z}{2}\)

\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{x}{6}=\frac{z}{10}\)

\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)

Đặt \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}=k(k\inℤ)\)

\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\)

\(\Leftrightarrow x\cdot y=6k\cdot15k=90\)

\(\Leftrightarrow90:k^2=90\Leftrightarrow k^2=1\Leftrightarrow k=\pm1\)

\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=15\\z=10\end{cases}}\)hay \(\hept{\begin{cases}x=-6\\y=-15\\z=-10\end{cases}}\)

Vậy \((x,y)\in(6,15);(-6,-15)\)

d, \(2x=3y=5z\Leftrightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\)

\(\Leftrightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

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

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

Vậy : \(\hept{\begin{cases}\frac{x}{15}=5\\\frac{y}{10}=5\\\frac{z}{6}=5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=75\\y=50\\z=30\end{cases}}\)

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:\(x=6;y=16;z=10\)

Vì \(\hept{\begin{cases}3x=5y\\2y=-3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{y}{3}\\\frac{y}{-3}=\frac{z}{2}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{-2}\end{cases}}\Rightarrow\frac{x}{5}=\frac{y}{3}=\frac{z}{-2}\)

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

\(\frac{x}{5}=\frac{y}{3}=\frac{z}{-2}=\frac{x+y-z}{5+3-\left(-2\right)}=\frac{2}{10}=\frac{1}{5}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{5}.5=1\\y=\frac{1}{5}.3=\frac{3}{5}\\z=\frac{1}{5}.\left(-2\right)=\frac{-2}{5}\end{cases}}\)

Ta có : 

\(3x=5y\Leftrightarrow\frac{x}{5}=\frac{y}{3}\)

\(2y=-3z\Leftrightarrow\frac{y}{3}=-\frac{z}{2}\)

Do đó : 

\(\frac{x}{5}=\frac{y}{3}=-\frac{z}{2}\)

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

\(\frac{x}{5}=\frac{y}{3}=-\frac{z}{2}=\frac{x+y-z}{5+3-2}=\frac{2}{6}=\frac{1}{3}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{1}{3}\Rightarrow x=\frac{5}{3}\\\frac{y}{3}=\frac{1}{3}\Rightarrow y=1\\-\frac{z}{2}=\frac{1}{3}\Rightarrow-\frac{2}{3}\end{cases}}\)

Vậy ...