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.

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

2x−3y/5=5y−2z/3=3z−5x/2=10x-15y/25=15y-6z/9=6z-10x/4=...+..+..../25+9+4=0/31=0

=> 2x=3y;  5y=2z ;  3z=5x => x/3=y/2; y/2=z/5

=> x/3=y/2 =z/5 = 12x/36=5y/10=3z/15= (12x+5y-3z)/31

      x/3 = 3y/6=2z/10 = (x-3y+2z)/7

=>  (12x+5y-3z)/ (x-3y+2z)=31/7

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

\(=\dfrac{\left(2x-3y\right)^3}{\left(2x-3y\right)^2}=2x-3y\)

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3

Ta có: \(\dfrac{x}{y}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{2}\)

Đặt \(\dfrac{x}{3}=\dfrac{y}{2}=k\)

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

Ta có: \(H=\dfrac{2x-3y}{x-5y}\)

\(=\dfrac{2\cdot3k-3\cdot2k}{3k-5\cdot2k}=\dfrac{6k-6k}{3k-10k}=0\)

Ta có: ⇔x3=y2⇔x3=y2

Đặt H=2x−3yx−5yH=2x−3yx−5y

Toán lớp 1 khó thế à 

Bài này quá dễ:vv

Ta có 3y-x=6

=> \(\left\{{}\begin{matrix}3y=6+x\\x=3y-6\end{matrix}\right.\)

Thay vào A ta có: \(A=\dfrac{x}{y-2}+\dfrac{2x-3y}{x-6}=\dfrac{3y-6}{y-2}+\dfrac{2x-6-x}{x-6}=\dfrac{3\left(y-2\right)}{y-2}+\dfrac{x-6}{x-6}=3+1=4\)Vậy khi 3y-x=6 thì A=4

x/y=1/2 ->y=2x

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

Thấy đúng xin k nha

\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{-2\left(-4k\right)-7k+5.3k}{2.\left(-4k\right)-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-5k}=-\dfrac{16}{5}\)

Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)

\(\Rightarrow x=-4k;y=-7k;z=3k\) (1)

Thay (1) vào A , ta được

\(A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{2\left(-4k\right)-3\left(-7k\right)-6.3k}\)

\(\Rightarrow A=\dfrac{8k+\left(-7k\right)+15k}{-8k+21k+\left(-18k\right)}\)

\(\Rightarrow A=\dfrac{k[8+\left(-7\right)+15]}{k[-8+21+\left(-18\right)]}\)

\(\Rightarrow A=\dfrac{16k}{-5k}\)

\(\Rightarrow A=\dfrac{16}{5}\)

Vậy \(A=\dfrac{16}{5}\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)

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

Ta có: \(E=\dfrac{3x^2+5y^2}{4x^2-y^2}\)

\(=\dfrac{3\cdot\left(2k\right)^2+5\cdot\left(3k\right)^2}{4\cdot\left(2k\right)^2-\left(3k\right)^2}=\dfrac{3\cdot4k^2+5\cdot9k^2}{4\cdot4k^2-9k^2}\)

\(=\dfrac{12k^2+45k^2}{16k^2-9k^2}=\dfrac{57k^2}{7k^2}=\dfrac{57}{7}\)