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)

a) VP=(a-b)²+4ab

        =a²-2ab+b²+4ab

        =a²+b²+2ab

        =(a+b)²=VT

Vậy (a+b)^2 = (a-b)^2 +4ab

b) VP=(a+b)²-4ab

        =a²+2ab+b²-4ab

        =a²-2ab+b²

        =(a-b)²=VT

Vậy (a-b)^2 = (a+b)^2 - 4ab

c)

VP=(ax-by)²+(ay+bx)²

=a²x²-2axby+b²y²+a²y²+2axby+b²x²

=a²x²+b²y²+a²y²+b²x²

=(a²x²+b²x²)+(b²y²+a²y²)

=x².(a²+b²)+y².(a²+b²)

=(a²+b²)(a²+y²)=VT

Vậy ( a^2 + b^2 ).(x^2 +y^2) = (ax - by)^2 +(ay+bx)^2

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.

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 \(\dfrac{a}{x}=\dfrac{b}{y}\)

Trả lời:

(a² + b² ) ( x² + y² ) - (ax + by )²

= a²x² + a²y² + b²x² + b²y² - [ ( ax )² + 2.ax.by + ( by )² ]

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

= a²x² + a²y² + b²x² + b²y² - a²x² - 2axby - b²y² 

= a²y² - 2axby + b²x²

= ( ay )² - 2aybx + ( bx )²

= ( ay - bx )²  (đpcm)

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