Lời giải:
\((a^2+b^2)(x^2+y^2)=(ax+by)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

\(\Leftrightarrow a^2y^2-2axby+b^2x^2=0\)

\(\Leftrightarrow (ay)^2-2(ay)(bx)+(bx)^2=0\)

\(\Leftrightarrow (ay-bx)^2=0\Rightarrow ay=bx\) (đpcm)

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=x^2\left(a^2+b^2\right)+y^2\left(a^2+b^2\right)\)

\(=a^2x^2+b^2x^2+a^2y^2+b^2y^2\)

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

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

\(\Leftrightarrow a^2x^2+b^2x^2+a^2y^2+b^2y^2=a^2y^2+2abxy+b^2y^2\)

\(\Leftrightarrow a^2x^2+b^2x^2=2abxy\)

\(\Leftrightarrow a^2x^2+b^2x^2-2abxy=0\)

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

\(\Leftrightarrow ax-bx=0\left(đpcm\right)\)

Ta có:

VT = (x² + y²)(a² + b²)

= x²a² + x²b² + y²a² + y²b²

= (a²x² + b²y² + 2axby) + (a²y² - 2aybx + b²x²)

= (ax + by)² + (ay - bx)²

=> VT = VP => đpcm

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

\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+b^2y^2+2axby\)

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

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

Vậy ta có điều phải chứng minh.

a) Ta có: \(\left(a+b\right)^2=4ab\)<=> \(a^2+b^2+2ab=4ab\)

                                               <=> \(a^2-2ab+b^2=0\)

                                                <=> \(\left(a-b\right)^2=0\)=> a=b (đpcm)

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

<=> \(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

<=> \(a^2y^2+b^2x^2-2axby=0\)

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

<=>ay=bx(đpcm)

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

\(\Leftrightarrow a^2x^2+2abxy+b^2y^2=a^2x^2+a^2y^2+x^2b^2+b^2y^2\)

\(\Leftrightarrow2abxy=a^2y^2+x^2b^2\)

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

\(\Leftrightarrow ay=xb\)

hay \(\dfrac{a}{x}=\dfrac{b}{y}\)

Ta có : (a^2 + b^2)(x^2 + y^2) = (ax + by)^2 

=> a^2x^2 + a^2y^2 +B^2x^2 + b^2y^2 = a^2x^2 + b^2y^2 + 2axby 

=> chuyển vế trái sang phải: a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 - a^2x^2 - b^2y^2 - 2axby = 0

=> a^2y^2 + b^2x^2 - 2axby = 0

=> (ax - by)^2 = 0

Chỉ khi ax = by thì (ax - by)^2 = 0 => ax = by.