a, \(3x=5y=7z=>\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)

áp dụng tính chất dãy tỉ số = nhau

\(=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y+z}{35+21+15}=\dfrac{10}{71}\)

\(=>\dfrac{x}{35}=\dfrac{10}{71}=>x=\dfrac{350}{71}\)

\(=>\dfrac{y}{21}=\dfrac{10}{71}=>y=\dfrac{210}{71}\)

\(=>\dfrac{z}{15}=\dfrac{10}{71}=>z=\dfrac{150}{71}\)

b, \(\)\(6x=5y=>\dfrac{x}{5}=\dfrac{y}{6}=>\dfrac{x}{20}=\dfrac{y}{24}\)

có \(7y=8z=>\dfrac{y}{8}=\dfrac{z}{7}=>\dfrac{y}{24}=\dfrac{z}{21}\)

\(=>\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}\)

áp dụng t/c dãy tỉ số = nhau

\(=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}=\dfrac{3x+2y+4z}{60+48+84}=\dfrac{12}{192}=\dfrac{1}{16}\)

\(=>\dfrac{3x}{60}=\dfrac{1}{16}=>x=1,25\)

\(=>\dfrac{2y}{48}=\dfrac{1}{16}=>y=1,5\)

\(=>\dfrac{4z}{84}=\dfrac{1}{16}=>z=1,3125\)

c, \(x:y:z=1:2:3=>\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)

\(=>x=\dfrac{y}{2},z=\dfrac{3y}{2}\)

thay x,z vào \(x^3+y^3+z^3=36=>\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)

\(=>y=2\)

\(=>x=\dfrac{y}{2}=\dfrac{2}{2}=1,z=\dfrac{3y}{2}=\dfrac{3.2}{2}=3\)

d, \(\dfrac{x}{2}=\dfrac{y}{3}=>x=\dfrac{2y}{3}\)

thay x vào \(3x^3+y^3=51=>3.\left(\dfrac{2y}{3}\right)^3+y^3=51=>y=3\)

\(=>x=\dfrac{2.3}{3}=2\)

c, từ đoạn này á

\(\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)

\(< =>\dfrac{y^3}{8}+\dfrac{8y^3}{8}+\dfrac{27y^3}{8}=36\)

\(=>\dfrac{36y^3}{8}=36=>36y^3=8.36=>y^3=8=>y=2\)

\(x-\frac{1}{2}=y-\frac{2}{3}=z-\frac{3}{4}\)va \(x-2y+3z=14\)

\(\frac{\Rightarrow\left(x-1\right)}{2}=\frac{\left(-2y+4\right)}{-6}=\frac{\left(3z-9\right)}{12}\)

\(=\frac{\left(x-1-2y+4+3z-9\right)}{\left(2-6+12\right)}\)

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

\(\frac{\Rightarrow\left(y-2\right)}{2}=-2\Leftrightarrow x-1=-4\Leftrightarrow x=-3\)

\(\Rightarrow\frac{\left(y-2\right)}{3}=-2\Leftrightarrow x-1=-4\Leftrightarrow x=-3\)

\(\Rightarrow\frac{\left(x-3\right)}{4}=-2\Leftrightarrow z-3=-8\Leftrightarrow z=-5\)

\(b)\)

Theo đề ra:

\(x:y:z=3:4:5\)

\(2x^2+2y^2-3z^2=-100\)

\(\Leftrightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

\(\Leftrightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}\)

\(\Leftrightarrow\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}\)

Áp dụng tính chất dãy tỷ số bằng nhau:

\(\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}=\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}=4\)

\(\Leftrightarrow\hept{\begin{cases}\frac{x}{3}=4\Leftrightarrow x=12\\\frac{y}{4}=4\Leftrightarrow y=16\\\frac{z}{5}=4\Leftrightarrow z=20\end{cases}}\)

\(x:y:z=3:5:\left(-2\right)\)

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

Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=-\dfrac{16}{4}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-4\right).3=-12\\y=\left(-4\right).5=-20\\z=\left(-4\right).\left(-2\right)=8\end{matrix}\right.\)

`x : y : z= 3:4:5`

`=> x/3 = y/4 = z/5 <=> x^2/9 = y^2/16 = z^2/25`

Áp dụng dãy tỉ số bằng nhau:

`x^2/9 = y^2/16 = z^2/25 = (2x^2 + 2y^2 - 3z^2)/(18 + 32 - 75) = -100/-25 = 4`.

`=> {(x^2/9 = 4 => x = +-6), (y^2/16 =4 <=> x = +-8), (z^2/25 = 4 => z = +-10):}`

Vậy ...

+) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)

=> \(\hept{\begin{cases}\frac{x^2}{9}=4\\\frac{y^2}{16}=4\end{cases}}\) => \(\hept{\begin{cases}x^2=4.9=36\\y^2=4.16=64\end{cases}}\) => \(\hept{\begin{cases}x=\pm6\\y=\pm8\end{cases}}\)

Vậy ...