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Ta có:\(\frac{15z-20y}{\frac{5}{12}}=\frac{12x-15z}{\frac{3}{20}}=\frac{20y-12x}{\frac{4}{15}}=0\)

=>3z-4y=0,

4x-5z=0,

5y-3x=0

=>3z=4y,

4x=5z,

5y=3x.

Rồi chuyển thành tỉ số và làm tiếp

Đổi thành \(\frac{3z-4y}{\frac{1}{12}}=\frac{4x-5z}{\frac{1}{20}}=\frac{5y-3x}{\frac{1}{15}}\)

Sau đó áp dụng dãy TSBN rút về x/a=y/b=z/t rồi làm tiếp

bạn ơi, hình như đề này sai rồi thì phải 

1,7y

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:

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

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

\(3x=2y;4y=5z\) 

\(\Rightarrow\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{15}=\frac{z}{12}\)

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\Rightarrow\)\(\frac{2x}{20}=\frac{3y}{45}=\frac{5z}{60}=\frac{2x-3y+5z}{125}=\frac{21}{125}\)

\(\frac{2x}{20}=\frac{21}{125}.....................\)

\(\frac{3y}{45}=\frac{21}{125}......................\)

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Giải thì giải cho hết luôn đi

a) 5y = 72

=> y = 72/5

2x = 3y

<=> 2x = 3 . 72/5

<=> 2x = 216 / 5

<=> x =108/5

3x - 7y + 5z = -30

<=> 3 . 108/5 - 7. 72/5 + 5z = - 30

<=> 324/5 - 504/5 +5z = -30

<=> 5z = 6

<=> x = 6/5 

câu a đoạn cuối z = 6/5 nha 

b) x : y : z = 5 : 3 :4 

\(\Leftrightarrow\frac{x}{5}=\frac{y}{3}=\frac{z}{4}\Leftrightarrow\frac{x}{5}=\frac{2y}{6}=\frac{z}{4}\)

Áp dụng t/c dãy tỉ số = nhau , ta có 

\(\frac{x}{5}=\frac{2y}{6}=\frac{z}{4}=\frac{x+2y-z}{5+6-4}=\frac{-121}{7}\)

=> x =-605/ 7

=> y = -363 / 7

=> z = -484 / 7

TA CÓ  \(\frac{3x-5y}{2}=\frac{7y-3z}{3}=\frac{5z-7x}{4}\)\(=\frac{21x-35y}{14}=\frac{35y-15z}{15}=\frac{15z-21x}{12}\)=\(\frac{21x-35+35y-15z+15z-21x}{14+15+12}=\frac{0}{41}=0\)

=> \(\hept{\begin{cases}3x-5y=0\\7y-3z=0\\5z-7x=0\end{cases}\left(=\right)\hept{\begin{cases}3x=5y\\7y=3z\\5z=7x\end{cases}\left(=\right)\hept{\begin{cases}\frac{x}{5}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{7}\\\frac{z}{7}=\frac{x}{5}\end{cases}}}}\)

=> \(\frac{x}{5}=\frac{y}{3}=\frac{z}{7}=\frac{x+y+z}{5+3+7}=\frac{17}{15}\)

=>\(\hept{\begin{cases}x=\frac{17}{3}\\y=\frac{17}{5}\\z=\frac{119}{15}\end{cases}}\)

ai trả lời được câu này mình cho 5 k

tìm x, biết

10+11+12+13+.....x=5106