Bài 1:
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

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

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

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

\(\Leftrightarrow ay=bx\)

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

\(\Rightarrowđpcm\)

Bài 2:

Ta có: \(VT=\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)

\(=\left(5a-3b\right)^2-64c^2\)

\(=25a^2-30ab+9b^2-64c^2\)

\(=25a^2-30ab+9b^2-16a^2+16b^2\left(a^2-b^2=4c^2\right)\)

\(=9a^2-30ab+25b^2=\left(3a-5b\right)^2=VP\)

\(\Rightarrowđpcm\)

hey Xuân phẹt oy là tui đây

bài 2

a²(b-c)+b²(c-a)+c²(a-b)=a²b-a²c+b²c-b²a+c²a-c²b=b(a²-c²)+ac(a-c)-b²(a-c)=(a-c)(ab-bc+ac-b²)=(a-c)(c-b)(a-b)=0

=>a-c=0 hoặc c-b=0 hoặc a-b=0

=>c=a hoặc c=b hoặc a=b

=>đpcm

nhớ tick vs nha

Ta có : \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)

\(=\left(5a-3b\right)^2-\left(8c\right)^2\)

\(=\left(5a-3b\right)^2-64c^2\)

\(=\left(5a-3b\right)^2-16.4c^2\)

\(=\left(5a-3b\right)^2-16\left(a^2-b^2\right)\)

\(=25a^2-30ab+9b^2-16a^2+16b^2\)

\(=9a^2-30ab+25b^2\)

\(=\left(3a-5b\right)^2\left(đpcm\right)\)

biến đổi vế trái 

\(\Leftrightarrow\left(5a-3b\right)^2-\left(8c\right)^2\)

\(\Leftrightarrow25a^2-30ab+9b^2-64c^2\)

\(\Leftrightarrow25a^2-30ab+9b^2-16\left(a^2-b^2\right)\)

\(\Leftrightarrow\left(25a^2-16a^2\right)-30ab+\left(9b^2+16b^2\right)\)

\(\Leftrightarrow9a^2-30ab+25b^2\)

\(\Leftrightarrow\left(3a-5b\right)^2\)  (điều cần c/m)