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nè, không làm thôi ằ nhagg. khó thì đừng gửi câu trả lời làm gì cho mệt nha bạn 

Ta có \(\frac{x-3}{y-2}=\frac{3}{2}\)

Nên 2 ( x - 3 ) = 3 ( y - 2 )

Do đó 2x - 6 = 3y - 6 nên 2x = 3y

Suy ra 2x - 2y = y hay 2 ( x - y ) = y

Mà x - y = 4 nên 2.4 = y

Suy ra y = 8

\(x=\frac{3y}{2}=\frac{3.8}{2}=12\)

Vây x = 12 ; y = 8

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

=> \(\left(x-3\right).2=3\left(y-2\right)\)

=> \(2x-6=3y-6\)

=> \(2x=3y\)

Khi đó ,ta có: x - y = 4

=> 2(x - y) = 8

=> 2x - 2y = 8

=> 3y - 2y = 8

=> y = 8 

     => x = 4 + 8 = 12

Vậy x = 12 và y = 8

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

=> 2. ( x - 3 ) = 3.( y - 2 )

=> 2x - 6 = 3y - 6

=> 2x  = 3y

=> \(\frac{x}{3}=\frac{y}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau.

=> \(\frac{x}{3}=\frac{y}{2}=\frac{x-y}{3-2}\)\(=\frac{4}{1}=4\)

=> \(\hept{\begin{cases}x=3.4=12\\y=2.4=8\end{cases}}\)

Vậy x = 8, y = 12

ta có:\(\frac{x-3}{y-2}=\frac{3}{2}\)

\(\Rightarrow\frac{x-3}{3}=\frac{y-2}{2}=\frac{x-3-y+2}{3-2}=\frac{\left(x-y\right)-\left(3-2\right)}{1}=\frac{4-1}{1}=3\)

=> x-3/3 = 3 => x-3 = 9 => x = 12

y-2/2 = 3 => y-2 = 6=> y = 8

Ta có: \(\frac{x-3}{y-2}=\frac{3}{2}\)

=> (x - 3) . 2 = (y - 2) . 3

=> 2x - 6 = 3y - 6

=> 2x = 3y

=> x = 3/2y

Khi đó, ta có: x - y = 4

hay 3/2y - y = 4

=> 1/2y = 4

=> y = 4 : 1/2

=> y = 8

=> x = 4 + 8 = 12

Vậy x = 12 và y = 8

Theo đề:\(\frac{x-3}{y-2}=\frac{3}{2}\Rightarrow2\left(x-3\right)=3\left(y-2\right)\)

\(\Rightarrow2x-6=3y-6\)

\(\Rightarrow2y-3y=-6+6\)

Vì\(x-y=4\Rightarrow x=4+y\)

\(\Rightarrow2\left(y+4\right)-3y=0\)

\(\Rightarrow2y+8-3y=0\)

\(\Rightarrow-y=-8\Rightarrow y=8\)

\(\Rightarrow x=y+4=8+4=12\)

Vậy y=8;x=12

Ta có :

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

\(\Rightarrow2\left(x-3\right)=3\left(y-2\right)\)

\(\Rightarrow2x-6=3y-6\)

\(\Rightarrow2x-3y=6-6\)

\(\Rightarrow2x-3y=0\)

Vì \(x-y=4\)\(\Rightarrow\)\(x=4+y\)

\(\Rightarrow2\left(4+y\right)-3y=0\)

\(\Rightarrow8+2y-3y=0\)

\(\Rightarrow8-y=0\)

\(\Rightarrow y=8\)

Lại có : \(x=4+y\)

       \(\Rightarrow x=4+8\)

       \(\Rightarrow x=12\)

Vậy x = 12 và y = 8

_Chúc bạn học tốt_

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

\(\Leftrightarrow\left(x-3\right)2=\left(y-2\right)3\)

\(\Leftrightarrow2x-6=3y-6\)

\(\Leftrightarrow\left(2x-6\right)-\left(3y-6\right)=0\)

\(\Leftrightarrow2x-6-3y+6=0\)

\(\Leftrightarrow\left(2x-3y\right)-\left(6+6\right)=0\)

\(\Leftrightarrow\left(2x-3y\right)-12=0\)

\(\Leftrightarrow2x-3y=12\)

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

\(\Rightarrow\frac{x}{3}=\frac{y}{2}=\frac{x-y}{3-2}=\frac{4}{1}=4\)

=>x=12;y=8

\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

=> x - 29 = 0

=> x = 29.