Ta có: x−y1=x+y7=(x−y)+(x+y)1+7=2x8=x4x−y1=x+y7=(x−y)+(x+y)1+7=2x8=x4

xy=xy24⇔6x24=xy24xy=xy24⇔6x24=xy24

⇒6x=xy⇒6x=xy

⇒y=6⇒y=6

x−61=x+67x−61=x+67

⇔7.(x−6)=x+6⇔7.(x−6)=x+6

⇔7x−42=x+6⇔7x−42=x+6

⇔7x−x=6+42⇔7x−x=6+42

⇔6x=48⇔6x=48

⇒x=8⇒x=8

Vậy x=8;y=6

ko can tl dau, mk biet lam rui may bn ak . Dung tl nha

Ta có : \(\frac{x-y}{1}=\frac{x+y}{7}=\frac{\left(x-y\right)+\left(x+y\right)}{1+7}=\frac{2x}{8}=\frac{x}{4}\)

\(\frac{x}{4}=\frac{xy}{24}\Leftrightarrow\frac{6x}{24}=\frac{xy}{24}\) => 6x = xy => y = 6

\(\frac{x-6}{1}=\frac{x+6}{7}\)

<=> 7(x - 6) = x + 6

<=> 7x - 42 = x + 6

<=> 7x - x = 6 + 42

<=> 6x = 48

=> x = 8

Vậy x = 8 ; y = 6

xy=48

48

xy=48

48

=> \(\frac{x-y}{1}=\frac{x+y}{7}=\frac{xy}{24}\)

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

\(\frac{xy}{24}=\frac{x-y}{1}=\frac{x+y}{7}=\frac{\left(x-y\right)+\left(x+y\right)}{1+7}=\frac{\left(x-y\right)-\left(x+y\right)}{1-7}\)=> \(\frac{xy}{24}=\frac{x}{4}=\frac{y}{3}\)

\(\frac{xy}{24}=\frac{x}{4}\)=>\(\frac{x}{4}.\frac{y}{6}=\frac{x}{4}\)=>  \(\frac{y}{6}=\frac{x}{4}:\frac{x}{4}=1\) ( do x khác 0) => y = 6

\(\frac{xy}{24}=\frac{y}{3}\Rightarrow\frac{x}{8}.\frac{y}{3}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{3}:\frac{y}{3}=1\) ( do y khác 0) => x = 8

Vậy...