4x(3y + 11)=6y(2x + 8)  =>12xy + 44x =12xy + 48y

=>44x = 48y  =>\(\frac{x}{y}\)=\(\frac{48}{44}\)

Ta dùng tỉ lệ thức bạn:

\(\frac{4x}{6y}=\frac{2x+8}{3y+11}\)

<=>4x(3y+11)=6y(2x+8)

<=>12xy+44x=12xy+48y

<=>48y-44x=0

<=>12y-11x=0

<=>12y=11x

Đến đây mình cũng chịu mà hình như đến đây là xong phải

\(\frac{4x}{6y}=\frac{2x+8}{3y+11}\)

<=>4x(3y+11)=6y(2x+8)

<=>12xy+44x=12xy+48y

<=>44x=48y(cùng bớt đi 12xy)

do đó x/y=48/44=12/11

\(\frac{4x}{6y}=\frac{2x+8}{3y+11}\)

Ta có:4x*3y+44x=6y*2x+48y

12xy+44x=12yx+48y

44x=48y

=>\(\frac{x}{y}=\frac{48}{44}\)

=>x/y=\(\frac{12}{11}\)

a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}\left(ĐK:x\ne0;x\ne-4\right)\)

\(=\frac{6}{x\left(x+4\right)}+\frac{3}{2\left(x+4\right)}=\frac{12+3x}{2x\left(x+4\right)}=\frac{3\left(4+x\right)}{2x\left(x+4\right)}=\frac{3}{2x}\)

b) \(\frac{4xy-5}{140x^3y}-\frac{6y^2-5}{10x^3y}\left(ĐK:x,y\ne0\right)\)

\(=\frac{4xy-5-6y^2+5}{10x^3y}=\frac{2y\left(2x-3y\right)}{10x^3y}=\frac{2x-3y}{5x^3}\)

từ trên

=>4x(3y+11)=6y(2x+8)

=>12xy+44=12xy+48

=>\(\frac{x}{y}=\frac{12}{11}\)

\(\frac{12}{11}\)

\(\frac{4x}{6y}=\frac{2x+8}{3y+11}\Leftrightarrow4x\left(3y+11\right)=6y\left(2x+8\right)\Leftrightarrow12xy+44x=12xy+48y\)

\(44x=48y\Leftrightarrow\frac{x}{y}=\frac{48}{44}=\frac{12}{11}\)

\(\frac{4x}{6y}=\frac{2x+8}{3y+11}\Leftrightarrow4x\left(3y+11\right)=6y\left(2x+8\right)\Leftrightarrow12xy+44x=12xy+48y\)

\(\Leftrightarrow44x=48y\Leftrightarrow\frac{x}{y}=\frac{48}{44}=\frac{12}{11}\)