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

\(\Leftrightarrow\frac{20z-24y}{4^2}=\frac{30x-20z}{5^2}=\frac{24y-30x}{6^2}\)

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

\(\frac{20z-24y}{4^2}=\frac{30x-20z}{5^2}=\frac{24y-30x}{6^2}=\frac{20z-24y+30x-20z+24y-30x}{4^2+5^2+6^2}\)

\(=\frac{0}{4^2+5^2+6^2}=0\)

\(\Rightarrow\hept{\begin{cases}20z=24y\\30x=20z\\24y=30x\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}5z=6y\\6x=4z\\4y=5x\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{z}{6}=\frac{y}{5}\\\frac{x}{4}=\frac{z}{6}\\\frac{y}{5}=\frac{x}{4}\end{cases}}\)

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

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

Sau đó, áp dụng tính chất của dãy tỉ số bằng nhau là được nhé.

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

\(\Leftrightarrow\dfrac{4\left(5z-6y\right)}{16}=\dfrac{5\left(6x-4z\right)}{25}=\dfrac{6\left(4y-5x\right)}{36}\)

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

ADTCDTSBN có:

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

Do đó \(20z-24y=0;30x-20z=0\)

\(\Leftrightarrow5z=6y;6x=4z\)

\(\Rightarrow y=\dfrac{5z}{6};x=\dfrac{4z}{6}\)

Có \(3x-3y+5z=96\Rightarrow3.\dfrac{4z}{6}-3.\dfrac{5z}{6}+5z=96\)

\(\Rightarrow z=\dfrac{64}{3}\) \(\Rightarrow y=\dfrac{160}{9}\)và \(x=\dfrac{128}{9}\)

Vậy...

cho x/3 = y/4 và y/5 = z/6. tìm M = 2x + 3y+ 4z / 3x + 4y + 5z

moi hok lop 6

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

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

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

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

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

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

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

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

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

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

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