\(\frac{1+7y}{7x}=\frac{1+9y}{2x}\) \(\Leftrightarrow\frac{1+7y}{7}=\frac{1+9y}{2}\)

\(\Leftrightarrow\left(1+7y\right)2=7\left(1+9y\right)\)

\(\Leftrightarrow2+14y=7+63y\)

\(\Leftrightarrow63y-14y=2-7\)

\(\Leftrightarrow y=-\frac{5}{49}\)

Thay \(x=-\frac{5}{49}\) vào biểu thức ta có :

\(\frac{1+7.\frac{-5}{49}}{7.x}=\frac{1+9.\frac{-5}{49}}{2x}\)

\(\Leftrightarrow x=2\)

Vậy..

Lời giải:

Ta có:
\(\frac{1+7y}{7x}=\frac{1+9y}{2x}\Rightarrow \frac{1+7y}{7}=\frac{1+9y}{2}\)

\(\Rightarrow 2(1+7y)=7(1+9y)\)

\(\Leftrightarrow 49y+5=0\Rightarrow y=\frac{-5}{49}\). Thay giá trị trên của $y$ vào điều kiện ban đầu ta có:

\(\frac{1+5y}{24}=\frac{1+9y}{2x}\)

\(\Leftrightarrow \frac{1+5.\frac{-5}{49}}{24}=\frac{1+9.\frac{-5}{49}}{2x}\)

\(\Leftrightarrow x=4\)

Vậy \(x=4; y=\frac{-5}{49}\)

Theo bài ra: \(\dfrac{1+5y}{24}=\dfrac{1+7y}{7x}=\dfrac{1+9y}{2x}\left(1\right)\)

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

\(\dfrac{1+5y}{24}=\dfrac{1+7y}{7x}=\dfrac{1+9y}{2x}=\dfrac{-2y}{24-7x}=\dfrac{-2y}{5x}\)

TH1: \(y=0\)

\(\Rightarrow\left(1\right)\Rightarrow\dfrac{1}{24}=\dfrac{1}{7x}=\dfrac{1}{2x}\) (vô lí)

\(\Rightarrow\) Loại

TH2: \(y\ne0\)

\(\Rightarrow\dfrac{-2y}{24-7x}=\dfrac{-2y}{5x}\)

\(\Rightarrow24-7x=5x\)

\(\Rightarrow12x=24\)

\(\Rightarrow x=2\)

Thay \(x=2\) vào \(\dfrac{1+5y}{24}=\dfrac{1+7y}{7x}\) , ta được:

\(\dfrac{1+5y}{24}=\dfrac{1+7y}{14}\)

\(\Rightarrow\left(1+5y\right)14=\left(1+7y\right)24\)

\(\Rightarrow14+70y=24+168y\)

\(\Rightarrow70y-168y=24-14\)

\(\Rightarrow-98y=10\)

\(\Rightarrow y=-\dfrac{5}{49}\)

Vậy \(x=2;y=-\dfrac{5}{49}\)

\(A=4x^2+12xy+9y^2\)

\(B=25x^2-10xy+y^2\)

\(C=8x^3+12x^2y^2+6xy^4+y^6\)

\(D=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4y^2}{25}\)

\(E=x^3-27y^3\)

\(F=x^6-27\)

ý bạn là \(1+\frac{5y}{24}\)hay là \(\frac{1+5y}{24}\)

Cái thứ 2 í

\(\dfrac{1+7y}{7x}=\dfrac{1+9y}{2x}\Rightarrow\dfrac{2+14y}{14x}=\dfrac{7+63y}{14x}\)

\(\Rightarrow2+14y=7+63y\Rightarrow49y=-5\Rightarrow y=\dfrac{-5}{49}\)

\(\Rightarrow\dfrac{1+5\left(\dfrac{-5}{49}\right)}{24}=\dfrac{1+7\left(\dfrac{-5}{49}\right)}{7x}\)

\(\Rightarrow\dfrac{1}{49}=\dfrac{2}{49x}\Rightarrow x=2\)

Vậy \(\left\{{}\begin{matrix}x=2\\y=\dfrac{-5}{49}\end{matrix}\right.\)