\(2x=3y=5z\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)(Cùng chia cho 30 )

Áp dụng t/c  dãy tsbn \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

\(\Rightarrow\hept{\begin{cases}x=5.15=75\\y=5.10=50\\z=5.6=30\end{cases}}\)

Ta có 2x=3y=5z

\(\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\)

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

\(\frac{x}{15}=5\Rightarrow x=75\)

\(\frac{y}{10}=5\Rightarrow y=50\)

\(\frac{z}{6}=5\Rightarrow z=30\)

Vậy x = 75 ; y = 50 ; z = 30

cảm ơn hai bạn

\(2x=3y\text{⇒}\dfrac{x}{3}=\dfrac{y}{2}\text{⇒}\dfrac{x}{15}=\dfrac{y}{10}\)

\(2y=5z\text{⇒}\dfrac{y}{5}=\dfrac{z}{2}\text{⇒}\dfrac{y}{10}=\dfrac{z}{4}\)

⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{4}\)

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

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{4}=\dfrac{\left|x+y+z\right|}{\left|15+10+4\right|}=\dfrac{29}{29}=1\)

⇒x=15;y=10;z=4

Từ \(2x=3y=5z\Rightarrow\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}\)

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

\(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{57}{\frac{19}{30}}=90\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{\frac{1}{2}}=90\Rightarrow x=90\cdot\frac{1}{2}=45\\\frac{y}{\frac{1}{3}}=90\Rightarrow y=90\cdot\frac{1}{3}=30\\\frac{z}{\frac{1}{5}}=90\Rightarrow z=90\cdot\frac{1}{5}=18\end{cases}}\)

Khi đó \(x^2-y^2+z^2=45^2-30^2+18^2=1449\)

a) 2x=3y=4z⇒\(\dfrac{2}{\dfrac{1}{x}}=\dfrac{3}{\dfrac{1}{y}}=\dfrac{4}{\dfrac{1}{z}}=\dfrac{2+3+4}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=\dfrac{9}{3}=3\) ( Vì\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\))

⇒ x=\(\dfrac{3}{2}\) ; y=1; z=\(\dfrac{3}{4}\)

b) \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-by}{c}\)

= \(\dfrac{abz-acy}{a^{2^{ }}}=\dfrac{bcx-abz}{b^{2^{ }}}=\dfrac{acy-bcy}{c^2}\) =\(\dfrac{\left(abz-acy\right)+\left(bcx-abz\right)+\left(acy-bcy\right)}{a^2+b^2+c^{2^{ }}}=\dfrac{0}{a^2+b^2+c^{2^{ }}}=0\)

⇒\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-by}{c}=0\)

⇒ ✽bz-cy=0⇒bz=cy⇒\(\dfrac{b}{y}=\dfrac{c}{z}\) (1)

✽ cx-az=0⇒cx=az⇒ \(\dfrac{a}{x}=\dfrac{c}{z}\) (2)

Từ (1) và (2) suy ra\(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

Bài 1:

Ta có: \(\frac{x}{4}=\frac{y}{3}.\)

=> \(\frac{3x}{12}=\frac{2y}{6}\) và \(3x-2y=15.\)

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

\(\frac{3x}{12}=\frac{2y}{6}=\frac{3x-2y}{12-6}=\frac{15}{6}=\frac{5}{2}.\)

\(\left\{{}\begin{matrix}\frac{x}{4}=\frac{5}{2}\Rightarrow x=\frac{5}{2}.4=10\\\frac{y}{3}=\frac{5}{2}\Rightarrow y=\frac{5}{2}.3=\frac{15}{2}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(10;\frac{15}{2}\right).\)

Chúc bạn học tốt!