Lời giải:
\(\frac{4x-5y}{3}=\frac{5z-3x}{4}=\frac{3y-4z}{5}\)

\(=\frac{3(4x-5y)}{9}=\frac{4(5z-3x)}{16}=\frac{5(3y-4z)}{25}\)

\(=\frac{12x-15y}{9}=\frac{20z-12x}{16}=\frac{15y-20z}{25}=\frac{12x-15y+20z-12x+15y-20z}{9+16+25}=0\)

\(\Rightarrow 4x-5y=5z-3x=3y-4z=0\)

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

Ta có: \(\frac{4x-5y}{3}=\frac{5z-3x}{4}=\frac{3y-4z}{5}.\)

\(\Rightarrow\frac{3.\left(4x-5y\right)}{3.3}=\frac{4.\left(5z-3x\right)}{4.4}=\frac{5.\left(3y-4z\right)}{5.5}\)

\(\Rightarrow\frac{12x-15y}{9}=\frac{20z-12x}{16}=\frac{15y-20z}{25}.\)

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

\(\frac{12x-15y}{9}=\frac{20z-12x}{6}=\frac{15y-20z}{25}=\frac{12x-15y+20z-12x+15y-20z}{9+16+25}=\frac{0}{50}=0\)

\(\left\{{}\begin{matrix}\frac{4x-5y}{3}=0\\\frac{5z-3x}{4}=0\\\frac{3y-4z}{5}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x-5y=0\\5z-3x=0\\3y-4z=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x=5y\\5z=3x\\3y=4z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{5}=\frac{y}{4}\\\frac{z}{3}=\frac{x}{5}\\\frac{y}{4}=\frac{z}{3}\end{matrix}\right.\Rightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\left(đpcm\right).\)

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

Chúc bạn học tốt!

mik mới lớp 5 =]]

nên ko biết =]]]]]

=]]]]]]]]

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

bài này của bạn chx đủ đk hay sao ý,xem lại đề đi

tra mạng í bn

Đặt x/3=y/4=z/5=k

=>x=3k; y=4k; z=5k

\(\dfrac{4x-3y}{2016}=\dfrac{4\cdot3k-3\cdot4k}{2016}=0\)

\(\dfrac{5y-4z}{2017}=\dfrac{5\cdot4k-4\cdot5k}{2017}=0\)

\(\dfrac{3z-5x}{2018}=\dfrac{3\cdot5k-5\cdot3k}{2018}=0\)

=>\(\dfrac{4x-3y}{2016}=\dfrac{5y-4z}{2017}=\dfrac{3z-5x}{2018}\)

1.

Có: \(\frac{4x-5y}{7}=\frac{5z-3x}{9}=\frac{3y-4z}{11}\\ \Leftrightarrow\frac{7}{7}.\left(\frac{4x-5y}{7}\right)=\frac{9}{9}.\left(\frac{5z-3x}{9}\right)=\frac{11}{11}.\left(\frac{3y-4z}{11}\right)\\ \Leftrightarrow\frac{28x-35y}{49}=\frac{45z-27x}{81}=\frac{33y-44z}{121}\)

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

\(\frac{28x-35y}{49}=\frac{45z-27x}{81}=\frac{33y-44z}{121}=\frac{28x-35y+45z-27x+33y-44z}{49+81+121}\)

tính ra nó đc x+ 2y +z ko đc tròn cho lắm..... mệt r tự nghĩ tiếp đi

1.

Ta có: \(\frac{4x-5y}{7}=\frac{5z-3x}{9}=\frac{3y-4z}{11}.\)

\(\Rightarrow\frac{7.\left(4x-5y\right)}{49}=\frac{9.\left(5z-3x\right)}{81}=\frac{11.\left(3y-4z\right)}{121}\)

\(\Rightarrow\frac{28x-35y}{49}=\frac{45z-27x}{81}=\frac{33y-44z}{121}.\)

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

\(\frac{28x-35y}{49}=\frac{45z-27x}{81}=\frac{33y-44z}{121}=\frac{28x-35y+45z-27x+33y-44z}{49+81+121}=\frac{\left(28x-27x\right)-\left(35y-33y\right)+\left(45z-44z\right)}{251}=\frac{x-2y+z}{251}.\)

Đoạn này chịu rồi.

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:

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

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