\(\frac{5x-2y}{3x+4y}=\frac{3}{4}\Rightarrow9x+12y=20x-8y\)

\(\Rightarrow11x=20y\)

\(\Rightarrow\frac{x}{y}=\frac{20}{11}\)

20/11 can cach lam kb nhac cho 

k nua nha

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

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

\(=>20x-8y=9x+12y\)

\(=>20x-9x=12y+8y\)

\(=>11x=20y\)

\(=>\frac{x}{y}=\frac{20}{11}\)

vay \(\frac{x}{y}=\frac{20}{11}\)

Ta dễ dàng thấy : \(y\ne0\)nên ta có thể chia hai số hạng của tỉ số \(\frac{5x-2y}{3x+4y}\)cho \(y\). Ta có:

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

Đặt \(\frac{x}{y}\)là t ta có: 4.(5t - 2) = 3.(3t + 4)

Giải đc t = \(\frac{20}{11}\)hay \(\frac{x}{y}=\frac{20}{11}\)

hok tốt

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

\(\Leftrightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)

\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)

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

\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12x-8y}{16}=0\\\dfrac{2z-4x}{3}=0\\\dfrac{4y-3z}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}12x-8y=0\\2x-4z=0\\4y-3z=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)

Từ giả thiết \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{4}\)

\(\Rightarrow\dfrac{3xz-2yz}{4z}=\dfrac{2yz-4xy}{3y}=\dfrac{4xy-3xz}{4x}\)

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

\(\dfrac{3xz-2yz}{4z}=\dfrac{2yz-4xy}{3y}=\dfrac{4xy-3xz}{4x}=\dfrac{3xz-2yz+2yz-4xy+4xy-3xz}{4z+3y+4x}=\dfrac{0}{4z+3y+4x}=0\)

\(\Rightarrow3xz=2yz=4xy\)

\(\Rightarrow\dfrac{3xyz}{y}=\dfrac{2xyz}{x}=\dfrac{4xyz}{z}\)

\(\Rightarrow\dfrac{3}{y}=\dfrac{2}{x}=\dfrac{4}{z}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)

Chúc bạn học tốt!

\(\dfrac{7x-3z}{5}=\dfrac{3y-5x}{7}=\dfrac{5z-7y}{3}\)

\(\Rightarrow\dfrac{35x-15z}{25}=\dfrac{21y-35x}{49}=\dfrac{15z-21y}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{35x-15z}{25}=\dfrac{21y-35x}{49}=\dfrac{15z-21y}{9}\)

\(=\dfrac{35x-15z+21y-35x+15z-21y}{25+49+9}\)

\(=\dfrac{0}{25+49+9}=0\)

\(\Rightarrow\left\{{}\begin{matrix}7x=3z\Rightarrow\dfrac{x}{3}=\dfrac{z}{7}\\3y=5x\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\\5z=7y\Rightarrow\dfrac{z}{7}=\dfrac{y}{5}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{3+5+7}=\dfrac{30}{15}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.5=10\\z=2.7=14\end{matrix}\right.\)

Tương tự

Xét \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\) (1)

Thay (1) vào P

=> P = \(\dfrac{3k+2.4k+3.5k}{2.5k+3.4k+4.5k}+\dfrac{2.5k+3.4k+4.5k}{3.3k+4.4k+5.5k}\) + \(\dfrac{3.3k+4.4k+5.5k}{4.3k+5.4k+6.5k}\)

=> P = \(\dfrac{26k}{42k}+\dfrac{42k}{50k}\) + \(\dfrac{50k}{62k}\)

=> P = \(\dfrac{13}{21}+\dfrac{21}{25}+\dfrac{25}{31}\approx2,265499232\)

lộn đề .

Thay 2z + 3y + 4z = 2x+ 3y + 4z nha

Lời giải:

 $\frac{x}{y}=\frac{2}{3}\Rightarrow \frac{x}{2}=\frac{y}{3}$. Đặt $\frac{x}{2}=\frac{y}{3}=k$ thì:

$x=2k; y=3k$

Khi đó: $3x-2y=3.2k-3.2k=0$. Mẫu số không thể bằng $0$ nên $A$ không xác định. Bạn xem lại.

$B=\frac{2(2k)^2-2k.3k+3(3k)^2}{3(2k)^2+2.2k.3k+(3k)^2}=\frac{29k^2}{33k^2}=\frac{29}{33}$

\(\frac{5x-2y}{3x+4y}=\frac{3}{4}\Rightarrow\left(5x-2y\right).4-3\left(3x+4y\right)=0\)

\(\Rightarrow20x-8y-9x-12y=0\)

\(\Rightarrow11x-20y=0\)

\(\Rightarrow11x=20y\)

\(\Rightarrow\frac{x}{y}=\frac{20}{11}\)

từ tỉ lệ thức đã cho ta có;

4(5x-2y)=3(3x+4y)

20x-8y=9x+12y

11x-8y=12y

11x=20y

x/y=20/11