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) \(\left(a+b\right)^2=a^2+2ab+b^2\left(1\right)\)

\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2-2ab+4ab+b^2=a^2+2ab+b^2\left(2\right)\)

Từ (1) và (2) => đpcm

b) \(\left(a-b\right)^2=a^2-2ab+b^2\left(3\right)\)

\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2+2ab-4ab+b^2=a^2-2ab+b^2\left(4\right)\)

Từ (3) và (4) =>đpcm

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

\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\left(5\right)\)

\(\left(ax-by\right)^2+\left(ay+bx\right)^2=a^2x^2-2axby+b^2y^2+a^2y^2+2aybx+b^2x^2\)

\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\left(6\right)\)

Từ (5) và (6) =>đpcm

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.

Bạn viết đề sai tứ tung luôn :v

Điều cần phải chứng minh:

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

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

\(VP=\left(ax-by\right)^2+\left(ay+bx\right)^2\)

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

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

\(VT=VP\rightarrowđpcm\)

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 \(\dfrac{a}{x}=\dfrac{b}{y}\)

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

\(\Leftrightarrow\) \(\dfrac{a}{x}=\dfrac{b}{y}\)

\(\Leftrightarrow ay=bx\)

\(\Leftrightarrow ay-bx=0\)

( Bất đẳng thức Bu - nhi - a - cốp - xki )