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

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

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

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

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

ai giúp mình với

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

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

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

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

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

\(\Rightarrow\hept{\begin{cases}\frac{4x-5y}{3}=0\\\frac{5z-3x}{4}=0\\\frac{3y-4z}{5}=0\end{cases}\Rightarrow\hept{\begin{cases}4x-5y=0\\5z-3x=0\\3y-4z=0\end{cases}\Rightarrow}\hept{\begin{cases}4x=5y\\5z=3x\\3y=4z\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{5}=\frac{y}{4}\\\frac{z}{3}=\frac{x}{5}\\\frac{y}{4}=\frac{z}{3}\end{cases}\Rightarrow}\frac{x}{5}=\frac{y}{4}=\frac{z}{3}}\)

Vậy...

Có gì không hiểu thì cứ hỏi nhé

Mình nghĩ đề phải là \(\frac{4x-5y}{3}=\frac{5z-3x}{4}=\frac{3y-4z}{5}\) chứ nhỉ???

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:

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

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

Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}\)

nên \(\dfrac{x}{15}=\dfrac{y}{20}\)(1)

Ta có: \(\dfrac{y}{5}=\dfrac{z}{6}\)

nên \(\dfrac{y}{20}=\dfrac{z}{24}\)(2)

Từ (1) và (2) suy ra \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)(3)

\(\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{4z}{96}\)

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

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{4z}{96}=\dfrac{2x+3y+4z}{30+60+96}=\dfrac{2x+3y+4z}{186}\)

Từ (3) suy ra \(\dfrac{3x}{45}=\dfrac{4y}{80}=\dfrac{5z}{120}\)

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

\(\dfrac{3x}{45}=\dfrac{4y}{80}=\dfrac{5z}{120}=\dfrac{3x+4y+5z}{45+80+120}=\dfrac{3x+4y+5z}{245}\)

Suy ra: \(M=\dfrac{2x+3y+4z}{3x+4y+5z}=\dfrac{186}{245}\)