Ta có: \(\frac{x}{a}=\frac{y}{b}\Leftrightarrow bx=ay\Leftrightarrow ay-bx=0\)

Ta có \(\left(x^2+y^2\right)\left(a^2+b^2\right)-\left(ax+by\right)^2\)

\(\Leftrightarrow\left(ax\right)^2+\left(bx\right)^2+\left(ay\right)^2+\left(by\right)^2-\left(ax\right)^2-2axby-\left(by\right)^2\)

\(\Leftrightarrow\left(ay\right)^2-2aybx+\left(bx\right)^2\)

\(\Leftrightarrow\left(ay-bx\right)^2=0\)

\(\Rightarrow\left(x^2+y^2\right)\left(a^2+b^2\right)-\left(ax+by\right)^2=0\)

\(\Leftrightarrow\left(x^2+y^2\right)\left(a^2+b^2\right)=\left(ax+by\right)^2\left(đpcm\right)\)

Ta có \(\left(x+y\right)^3\)=\(x^3+3x^2y+3xy^2+y^3\)

Mà \(x\left(x-3y\right)^2+y\left(y-3x\right)^2\)=\(x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\)

\(x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y\)\(=x^3+\left(-6x^2y+9x^2y\right)+\left(-6xy^2+9xy^2\right)+y^3\)

=\(x^3+3x^2y+3xy^2+y^3\)=\(\left(x+y\right)^3\)

=>đpcm

dạ e k gõ lm ak

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

áp dụng hằng đẳng thức thứ nhất ta có

\(^{\left(x+y\right)^2-y^2}\)=\(x^2+2xy+y^2-y^2\)

=\(x^2+2xy\)=x(x+2y) =>đpcm