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

=> (5x - 2y).4 = 7.(x + 3y)

=> 20x - 8y = 7x + 21y

=>> 20x - 7x = 21y + 8y

=> 13x = 29y

\(\Rightarrow\frac{x}{y}=\frac{29}{13}\)

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

\(\Rightarrow4\left(5x-2y\right)=7\left(x+3y\right)\)

\(\Rightarrow20x-8y=7x+21y\)

\(\Rightarrow20x-7x=8y+21y\)

\(\Rightarrow13x=29y\)

\(\Rightarrow\frac{x}{y}=\frac{29}{13}\)

Vậy \(\frac{x}{y}=\frac{29}{13}\)

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

\(\Rightarrow7\left(x+3y\right)=4\left(5x-2y\right)\)

\(\Rightarrow7x+21y=20x-8y\)

\(\Rightarrow7x-20x=-8y-21y\)

\(\Rightarrow-13x=-29y\)

\(\Rightarrow13x=29y\)

\(\Rightarrow\frac{x}{y}=\frac{29}{13}\)

Vậy : \(\frac{x}{y}=\frac{29}{13}\)

\(\frac{5x-2y}{x+3y}=chong.copy.linhtinh\Leftrightarrow20\left(\frac{x}{y}\right)-8=7\left(\frac{x}{y}\right)+21\Rightarrow13\left(\frac{x}{y}\right)=29\)

 \(\Rightarrow copy.linhtinh=bieuthuc\)không hiểu nhận được qua tin nhắn (hiểu rồi thì càng tốt)

\(\frac{x}{y}=\frac{29}{13}\)

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

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

\(20x-8y=7x+21y\)

\(13x=29y\)

\(\frac{x}{y}=\frac{29}{13}\)

Theo đế bài ta có:

\(20x-8y=7x+21y\)

\(\Leftrightarrow20x-7x=21y+8y\)

\(13x=21y\)

\(\Rightarrow\frac{x}{y}=\frac{21}{13}\)

a) Ta có:

\(\begin{array}{l}\frac{6}{{10}} = \frac{{6:2}}{{10:2}} = \frac{3}{5};\\\frac{9}{{15}} = \frac{{9:3}}{{15:3}} = \frac{3}{5}\end{array}\)

\(\begin{array}{l}\frac{{6 + 9}}{{10 + 15}} = \frac{{15}}{{25}} = \frac{{15:5}}{{25:5}} = \frac{3}{5};\\\frac{{6 - 9}}{{10 - 15}} = \frac{{ - 3}}{{ - 5}} = \frac{3}{5}\end{array}\)

Ta được: \(\frac{{6 + 9}}{{10 + 15}} = \frac{{6 - 9}}{{10 - 15}} = \frac{6}{{10}} = \frac{9}{{15}}\)

b) - Vì \(k = \frac{a}{b} \Rightarrow a = k.b\)

Vì \(k = \frac{c}{d} \Rightarrow c = k.d\)

- Ta có:

\(\begin{array}{l}\frac{{a + c}}{{b + d}} = \frac{{k.b + k.d}}{{b + d}} = \frac{{k.(b + d)}}{{b + d}} = k;\\\frac{{a - c}}{{b - d}} = \frac{{k.b - k.d}}{{b - d}} = \frac{{k.(b - d)}}{{b - d}} = k\end{array}\)

- Như vậy, \(\frac{{a + c}}{{b + d}}\) =\(\frac{{a - c}}{{b - d}}\) = \(\frac{a}{b}\) =\(\frac{c}{d}\)( = k)

a: \(\dfrac{6+9}{10+15}=\dfrac{15}{25}=\dfrac{3}{5};\dfrac{6-9}{10-15}=\dfrac{-3}{-5}=\dfrac{3}{5}\)

=>Bằng nhau

b: a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k;\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)

=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a}{b}=\dfrac{c}{d}\)

b) Áp dụng t/c dãy tỉ số bằng nhau,ta có:

\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{1+3y+1+5y+1+7y}{12+5x+4x}=\frac{3+15y}{12+5x+4x}=\frac{3\left(1+5y\right)}{2.3.2+5x+4x}=\frac{1+5y}{4+9x}=\frac{1+5y}{5x}\)<=> 4 + 9x = 5x

....

a/ Từ giả thiêt ta có \(\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\Leftrightarrow\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\)

\(\Leftrightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{40}\). Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{40}=k\)

\(\Rightarrow\begin{cases}x=15k\\y=20k\\z=40k\end{cases}\)

Theo đề bài : \(xy=1200\Leftrightarrow15k.20k=1200\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)

Tới đây dễ rồi nhé :)

b/ \(\frac{1+5y}{5x}=\frac{1+7y}{4x}\Leftrightarrow\frac{1+5y}{5}=\frac{1+7y}{4}\Leftrightarrow\frac{7+35y}{35}=\frac{5+35y}{20}=\frac{7+35y-5-35y}{35-20}=\frac{2}{15}\)

\(\Rightarrow y=-\frac{1}{15}\)

Thay y vào \(\frac{1+3y}{12}=\frac{1+5y}{5x}\) tìm được x = 2

Bài a) :)) Làm không ra. Tính mãi x.y vẫn không bằng 1200

Ta có: \(\frac{x+2y}{4x-3y}=-2\)

\(\Leftrightarrow x+2y=-2\left(4x-3y\right)\)

\(\Leftrightarrow x+2y=-8x+6y\)

\(\Leftrightarrow x+2y+8x-6y=0\)

\(\Leftrightarrow9x-4y=0\)

\(\Leftrightarrow9x=4y\)

hay \(\frac{x}{y}=\frac{4}{9}\)

Vậy: \(\frac{x}{y}=\frac{4}{9}\)