Ta có: \(x\left(x+2y\right)^3-y\left(y+2x\right)^3=27\)

\(\Leftrightarrow x\left(x^3+6x^2y+12xy^2+8y^3\right)-y\left(y^3+6xy^2+12x^2y+8x^3\right)=27\)

\(\Leftrightarrow x^4+6x^3y+12x^2y^2+8xy^3-y^4-6xy^3-12x^2y^2-8x^3y=27\)

\(\Leftrightarrow\left(x^4-y^4\right)-2x^3y+2xy^3=27\)

\(\Leftrightarrow\left(x^2-y^2\right)\left(x^2+y^2\right)-2xy\left(x^2-y^2\right)=27\)

\(\Leftrightarrow\left(x^2-y^2\right)\left(x^2-2xy+y^2\right)=27\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)^3=27\)

Vì x , y > 0 => \(x+y>0\Rightarrow\left(x-y\right)^3>0\Rightarrow x>y\)

Khi đó: \(\left(x-y\right)^3\in\left\{1;8;27\right\}\Rightarrow x-y\in\left\{1;2;3\right\}\)

Nếu \(\left(x-y\right)^3=1\Rightarrow\hept{\begin{cases}x-y=1\\x+y=27\end{cases}}\Rightarrow\hept{\begin{cases}x=14\\y=13\end{cases}}\)

Nếu \(\left(x-y\right)^3=8\Rightarrow\hept{\begin{cases}x-y=2\\x+y=\frac{27}{8}\end{cases}\left(ktm\right)}\)

Nếu \(\left(x-y\right)^3=27\Rightarrow\hept{\begin{cases}x-y=3\\x+y=1\end{cases}}\left(ktm\right)\)

Vậy x = 14 , y = 13

Từ định nghĩa bằng nhau của hai số phức, ta có:

a) ⇔ ;

b) ⇔ ;

c) ⇔ ⇔ .

2x(2y-14)-8(y-7)=0

=>\(4x\left(y-7\right)-8\left(y-7\right)=0\)

=>\(\left(y-7\right)\left(4x-8\right)=0\)

=>\(\left\{{}\begin{matrix}y-7=0\\4x-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7\\x=2\end{matrix}\right.\)

 4x (y - 7) - 8(y-7) =0

(4x-8) (y-7)=0
x=2 y =7

5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)

Thay từng TH rồi làm nha bạn

3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)

thay nhá

Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)

PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)

+) Với y = x - 1 thay vào pt (2):

\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))

Anh quy đồng lên đê, chắc cần vài con trâu đó:))

+) Với y = 2x + 3...

a

a

a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)

=3(x-y)^2+9

b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)

c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)