Ta có : \(\frac{5z-6y}{4}=\frac{6x-4z}{5}=\)\(\frac{4y-5x}{6}\)\(=\frac{20z-24y}{16}=\frac{30x-20z}{25}=\frac{24y-30x}{36}\)

\(=\frac{20z-24y+30x-20z+24y-30x}{16+25+36}\)\(=0\)

\(\Rightarrow\frac{5z-6y}{4}=0\Leftrightarrow5z-6y=0\Leftrightarrow5z=6y\Leftrightarrow\frac{y}{5}=\frac{z}{6}\left(1\right)\)

\(\Rightarrow\frac{6x-4z}{5}=0\Leftrightarrow6x-4z=0\Leftrightarrow6x=4z\Leftrightarrow\frac{z}{6}=\frac{x}{4}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\)\(\frac{y}{5}=\frac{z}{6}=\frac{x}{4}\)\(=\frac{3x}{12}=\frac{2y}{10}=\frac{5z}{30}\)

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

\(\frac{3x}{12}=\frac{2y}{10}=\frac{5z}{30}=\frac{3x-2y+5z}{12-10+30}=\frac{96}{32}=3\)

\(\Rightarrow\frac{x}{4}=3\Leftrightarrow x=3.4=12\)

\(\Rightarrow\frac{y}{5}=3\Leftrightarrow y=3.5=15\)

\(\Rightarrow\frac{z}{6}=3\Leftrightarrow z=3.6=18\)

Vậy \(\hept{\begin{cases}x=12\\y=15\\z=18\end{cases}}\)

       Bài làm :

Ta có :

 \(\frac{5z-6y}{4}=\frac{6x-4z}{5}=\frac{4y-5x}{6}=\frac{20z-24y}{16}=\frac{30x-20z}{25}=\frac{24y-30x}{36}\)\(\)

\(=\frac{20z-24y+30x-20z+24y-30x}{16+25+36}\)

\(=0\)

\(\Rightarrow\frac{5z-6y}{4}=0\Leftrightarrow5z-6y=0\Leftrightarrow5z=6y\Leftrightarrow\frac{y}{5}=\frac{z}{6}\left(1\right)\)

\(\Rightarrow\frac{6x-4z}{5}=0\Leftrightarrow6x-4z=0\Leftrightarrow6x=4z\Leftrightarrow\frac{z}{6}=\frac{x}{4}\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow\frac{y}{5}=\frac{z}{6}=\frac{x}{4}=\frac{3x}{12}=\frac{2y}{10}=\frac{5z}{30}\)

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

\(\frac{3x}{12}=\frac{2y}{10}=\frac{5z}{30}=\frac{3x-2y+5z}{12-10+30}=\frac{96}{32}=3\)

\(\Rightarrow\frac{x}{4}=3\Leftrightarrow x=3.4=12\)

\(\Rightarrow\frac{y}{5}=3\Leftrightarrow y=3.5=15\)

\(\Rightarrow\frac{z}{6}=3\Leftrightarrow z=3.6=18\)

Vậy x=12 ; y=15 ; z=18

Con tham khảo bài tương tự tại đây nhé:

Câu hỏi của ngoc Ngoc - Toán lớp 7 - Học toán với OnlineMath

\(\frac{5z-6y}{4}\)=\(\frac{6x-4y}{5}=\frac{4y-5x}{6}=\frac{20z-24y}{16}=\frac{30x-20z}{25}=\frac{24y-30x}{36}\)

=\(\frac{20z-24y+30x-20z+24y-30x}{16+25+36}=\frac{0}{77}=0\)

=>\(\frac{5z-6y}{4}=0=>5z-6y=0=>5z=6y=>\frac{y}{5}=\frac{z}{6}\left(1\right)\)

Tương tự ta có:\(\frac{x}{4}=\frac{y}{5}=\frac{z}{6}=\frac{3x}{12}=\frac{2y}{10}=\frac{5z}{30}\)

*Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{3z}{12}=\frac{2y}{10}=\frac{5z}{30}=\frac{3z-2y+5z}{12-10+30}=\frac{96}{32}=3\)

Tự giải nhé Đô Long

                                                              Kí tên: BTS V

moi hok lop 6

mình viết nhầm nha ở chỗ cuối là: 3x-2y+5z=96 nha

\(\frac{5z-6y}{4}=\frac{6x-4z}{5}=\frac{4y-5x}{6}\)

\(=\frac{4}{4}.\frac{5z-6y}{4}=\frac{5}{5}.\frac{6x-4z}{5}=\frac{6}{6}.\frac{4y-5x}{6}\)

\(=\frac{20z-24y}{16}=\frac{30x-20z}{25}=\frac{24y-30x}{36}=\frac{\left(20z-24y\right)+\left(30x-20z\right)+\left(24y-30x\right)}{16+25+36}\)

\(=\frac{20z-24y+30x-20z+24y-30x}{77}=\frac{\left(30x-30x\right)\left(24y-24y\right)\left(20z-20z\right)}{77}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{5z-6y}{4}=0\\\frac{6x-4z}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}5z=6y\\6x=4z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{z}{6}=\frac{y}{5}\\\frac{x}{4}=\frac{z}{6}\end{matrix}\right.\Rightarrow\frac{x}{4}=\frac{y}{5}=\frac{z}{6}\)

\(=\frac{3x}{12}=\frac{2y}{10}=\frac{5z}{30}=\frac{3x-2y+5z}{12-10+30}=\frac{96}{32}=3\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{4}=3\\\frac{y}{5}=3\\\frac{z}{6}=3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=3.4=12\\y=3.5=15\\z=3.6=18\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=12\\y=15\\z=18\end{matrix}\right.\)