5x - 2y/3x + 4y = 3/4

=> (5x - 2y) × 4 = (3x + 4y) × 3

=> 20x - 8y = 9x + 12y

=> 20x - 9x = 12y + 8y

=> 11x = 20y

=> x/y = 20/11

từ tỉ lệ thức ta có

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

20x-8y=9x+12y

11x-8y=12y

11x=20y

x/y=20/11

bài 1

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

\(\Rightarrow\left(3x-y\right).4=\left(x+y\right)3\)

\(\Rightarrow12x-4y=3x+3y\)

\(\Rightarrow9x=7y\)

\(\Rightarrow\frac{x}{y}=\frac{7}{9}\)

Vậy \(\frac{x}{y}=\frac{7}{9}\)

bài 2

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

\(\Rightarrow\) \(\left(3x+1\right)\left(5x+4\right)=\left(3x-1\right)\left(5x+7\right)\)

\(\Rightarrow15x^2+12x+10x+8=15x^2+21x-5x-7\)

\(\Rightarrow22x+8=16x-7\)

\(\Rightarrow22x+16x=-7-8\)

\(\Rightarrow6x=-15\)

\(\Rightarrow x=-2,5\)

Vậy x=-2,5

bạn giải đi bạn

Theo đề ta có: \(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

Đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\inℕ^∗\right)\)

Suy ra: \(x=3k;y=4k;z=5k\) Thay vào biểu thức P ta có:

\(P=\frac{3k+8k+15k}{6k+12k+20k}+\frac{6k+12k+20k}{9k+16k+25k}+\frac{9k+16k+25k}{12k+20k+30k}\)

\(P=\frac{26k}{38k}+\frac{38k}{50k}+\frac{50k}{62k}=\frac{13}{19}+\frac{19}{25}+\frac{25}{31}=\frac{33141}{14725}\)

Rút gọn A trước khi tính :

\(A=\left(\frac{7}{2}x^4y^3-\frac{1}{3}x^4y^3\right)+\left(8x^2y^5-5x^2y^5\right)-\left(6y+\frac{1}{2}y\right)\)

\(=\frac{19}{6}x^4y^3+3x^2y^5-\frac{13}{2}y\)

Thay \(x=-2,y=\frac{3}{4}\) vào A có :

\(A=\frac{19}{6}\cdot\left(-2\right)^4\cdot\left(\frac{3}{4}\right)^3+3\cdot\left(-2\right)^2\cdot\left(\frac{3}{4}\right)^5-\frac{13}{2}\cdot\frac{3}{4}\)

\(=\frac{171}{8}+\frac{729}{8192}-\frac{39}{8}\approx16,6\)

:)) Số xấu ....

Xét biểu thức A, ta suy ra:

\(A=\frac{19}{6}x^4y^3+3x^2y^5-\frac{-13}{2}y\)

Tại x=-2 và y=3/4 thì:

\(A=\frac{19}{6}\cdot\left(-2\right)^4\cdot\left(\frac{3}{4}\right)^3+3\cdot\left(-2\right)^2\cdot\left(\frac{3}{4}\right)^5-\frac{-13}{2}\cdot\frac{3}{4}\)

(phần này bạn tự tính)

\(\)

cảm ơn các bạn

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

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

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

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

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

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

\(13x=29y\)

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

\(\Rightarrow12x+20y=x-2y\Leftrightarrow11x=-22y\Rightarrow\frac{x}{y}=-\frac{22}{11}=-2\)

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

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

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

\(\Rightarrow11x=20y\)

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

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 \(\frac{x}{y}=\frac{20}{11}\)