dễ lắm nhưng bây h mình k có thời gian để giải 

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

a./ \(\frac{x}{5}=\frac{y}{4}=\frac{z}{7}=\frac{2y}{8}=\frac{x+2y+z}{5+8+7}=\frac{10}{20}=\frac{1}{2}\)

\(\Rightarrow x=\frac{5}{2};y=2;z=\frac{7}{2}\)

b./ \(\frac{x}{4}=\frac{y}{5}=\frac{z}{2}=\frac{x+y}{9}=\frac{18}{9}=2\)

\(\Rightarrow x=2\cdot4=8;y=2\cdot5=10;z=2\cdot2=4\)

a

Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)

\(\Rightarrow x=2k+1;y=3k+2;z=4k+3\)

Thay vào,ta được:

\(2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=50\)

\(\Leftrightarrow4k+2+9k+6-4k-3=50\)

\(\Leftrightarrow9k+5=50\)

\(\Leftrightarrow9k=45\)

\(\Leftrightarrow k=5\)

\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=\frac{5x-5}{10}=\frac{3y+9}{12}=\frac{4z-20}{24}\)

\(=\frac{5x-5-3y-9-4z+20}{10-12-24}=\frac{\left(5x-3y-4z\right)+\left(20-5-9\right)}{26}=\frac{46+6}{26}=2\)

\(\Rightarrow x=2\cdot2+1=5\)

\(y=4\cdot2-3=5\)

\(z=2\cdot6+5=17\)

Câu c tương tự như câu 1

\(c,\frac{2x}{5}=\frac{3y}{10}=\frac{z}{12}\)và x + y + z = 107

Ta có : \(\frac{2x}{5}=\frac{3y}{10}=\frac{z}{12}\Leftrightarrow\frac{x}{\frac{5}{2}}=\frac{y}{\frac{10}{3}}=\frac{z}{12}=\frac{x+y+z}{\frac{5}{2}+\frac{10}{3}+12}=\frac{107}{\frac{107}{6}}=107\cdot\frac{6}{107}=6\)

Vậy : \(\hept{\begin{cases}\frac{2x}{5}=6\\\frac{3y}{10}=6\\\frac{z}{12}=6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=15\\x=20\\z=72\end{cases}}\)

\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)

\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)

\(\hept{\begin{cases}\frac{x}{2}=\frac{x}{3}\\\frac{y}{5}=\frac{x}{7}\end{cases}\Rightarrow}\frac{x}{2}=\frac{5y}{15};\frac{3y}{15}=\frac{z}{7}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

Áp dụng tính chát dãy tỉ số = nhau ta có:

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)

\(\Rightarrow\frac{x}{10}=2\Rightarrow x=20\)

\(\frac{y}{15}=2\Rightarrow y=30\)

\(\frac{z}{21}=3\Rightarrow z=63\)

b, Tự làm

c, \(5x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{5}\)

\(2x=3z\Leftrightarrow\frac{x}{3}=\frac{z}{2}\)

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

\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{x}{6}=\frac{z}{10}\)

\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)

Đặt \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}=k(k\inℤ)\)

\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\)

\(\Leftrightarrow x\cdot y=6k\cdot15k=90\)

\(\Leftrightarrow90:k^2=90\Leftrightarrow k^2=1\Leftrightarrow k=\pm1\)

\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=15\\z=10\end{cases}}\)hay \(\hept{\begin{cases}x=-6\\y=-15\\z=-10\end{cases}}\)

Vậy \((x,y)\in(6,15);(-6,-15)\)

d, \(2x=3y=5z\Leftrightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\)

\(\Leftrightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

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

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

Vậy : \(\hept{\begin{cases}\frac{x}{15}=5\\\frac{y}{10}=5\\\frac{z}{6}=5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=75\\y=50\\z=30\end{cases}}\)

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.

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