Viết lại thành : \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Dựa theo tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

-> x = \(12.\dfrac{3}{2}=18\)

y =\(12.\dfrac{4}{3}=16\)

z =\(12.\dfrac{5}{4}\) = 15

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{z+4}{9}=\frac{2x+3y-1}{6x}\)(1)

Áp dụng tính chất dãy tỉ sổ bằng nhau, ta được

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{z+4}{9}=\frac{2x+3y-1}{6x}=\frac{\left(2x+1\right)+\left(3y-2\right)}{5+7}=\frac{2x+3y-1}{12}\)

\(\Rightarrow\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)

\(\Rightarrow\frac{12}{6x}=\frac{2x+3y-1}{2x+3y-1}=1\)

\(\Rightarrow\frac{2}{x}=1\)

\(\Rightarrow x=2\)

Thay x=2 vào (1), ta được

\(\frac{3y-2}{7}=\frac{z+4}{9}=\frac{2\cdot2+1}{5}=1\)

\(\Rightarrow\hept{\begin{cases}3y-2=7\\z+4=9\end{cases}}\Rightarrow\hept{\begin{cases}3y=9\\z=5\end{cases}}\Rightarrow\hept{\begin{cases}y=3\\z=5\end{cases}}\)

Vậy...hok tốt

Ta có : \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)

Nên : \(\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)

<=> 6x = 12

=> x = 2 . 

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}=\frac{\left(2x+1\right)+\left(3y-2\right)}{5+7}=\frac{2x+3y-1}{12}\)

\(\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)

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

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)

\(\Rightarrow\frac{2x+1}{5}=k\rightarrow2x+1=5k\rightarrow2k=5k-1\)

\(\frac{3y-2}{7}=k\rightarrow3y-2=7k\rightarrow3y=2k+2\)

 \(\frac{2x+3y-1}{6x}=k\rightarrow2x+3y-1=6x.k\)

                                     \(\rightarrow5k-1+7k+2-1=k.3\left(5k-1\right)\)

                                     \(\rightarrow12k=15k^2-3k\)

                                      \(\rightarrow15k^2-15k=0\)

                                       \(\rightarrow15k\left(k-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}k=0\rightarrow x=\frac{-1}{2};y=\frac{2}{3}\\k=1\rightarrow x=2;y=3\end{cases}}\)

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

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}=\frac{2x+3y+1-2}{5+7}=\frac{2x+3y-1}{12}\)

\(\Rightarrow\frac{2x+3y-1}{12}=\frac{2x+3y-1}{6x}\)

TH 1 : \(2x+3y-1=0\)

\(\Rightarrow\frac{2x+1}{5}=0;\frac{3y-2}{7}=0\)

\(\Rightarrow2x+1=0;3y-2=0\)

\(\Rightarrow2x=-1;3y=2\)

\(\Rightarrow x=-\frac{1}{2};y=\frac{2}{3}\)

TH 2 : \(2x+3y-1\ne0\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)

Mà \(\frac{2x+1}{5}=\frac{3y-2}{7}\)

\(\Rightarrow\frac{2.2+1}{5}=\frac{3y-2}{7}\)

\(\Rightarrow1=\frac{3y-2}{7}\)

\(\Rightarrow3y-2=7\)

\(\Rightarrow3y=9\)

\(\Rightarrow y=3\)

Vậy \(\orbr{\begin{cases}x=-\frac{1}{2};y=\frac{2}{3}\\x=2;y=3\end{cases}}\)

Theo t/c dãy tỉ số bằng nhau :

\(\Rightarrow\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+1+3y-2}{5+7}=\frac{2x+3y-1}{12}\)

Do \(\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)

\(\Rightarrow6x=12\Leftrightarrow x=2\)

Xét :\(\frac{2x+1}{5}=\frac{3y-2}{7}\)

\(1=\frac{3y-2}{7}\)

\(\Rightarrow3y=9\Leftrightarrow y=3\)

ta có: \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+1+3y-2}{5+7}=\frac{2x+3y-1}{12}\)

\(\Rightarrow\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\)

=> 6x = 12

x = 2

=>  \(\frac{2x+1}{5}=\frac{2.2+1}{5}=\frac{5}{5}=1\)

\(\frac{3y-2}{7}=1\Rightarrow3y-2=7\Rightarrow3y=9\Rightarrow y=3\)

KL: x = 2; y = 3

x=2

y=3

Áp dụng TC DCTSBN ta có :

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{\left(2x+1\right)+\left(3y-2\right)}{5+7}=\frac{2x+3y-1}{12}=\frac{2x+3y-1}{6x}\)

\(\Rightarrow6x=12\Rightarrow x=2\)

Thay x = 2 và 2 TLT đầu ta được :

\(\frac{2.2+1}{5}=\frac{3y-2}{7}\)

\(\Leftrightarrow\frac{3y-2}{7}=1\)

\(\Rightarrow3y-2=7\Rightarrow y=3\)

Vậy x = 2 và y = 3

Ta có: \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\) \(\left(x\ne0\right)\)

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

\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)\(=\frac{\left(2x+1\right)+\left(3y-2\right)-\left(2x+3y-1\right)}{5+7-6x}\)\(=\frac{0}{12-6x}=0\)

\(\Rightarrow\hept{\begin{cases}2x+1=0\\3y-2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{2}{3}\end{cases}}\)