Ta có: \(\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)

\(=\left(a^2c^2+2abcd+b^2d^2\right)+\left(a^2d^2-2abcd+b^2c^2\right)\)

\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

=> đpcm

( a² + b² )( c² + d² )

= a²c² + a²d² + b²c² + b²d²

= ( a²c² + 2abcd + b²d² ) + ( a²d² - 2abcd + b²c² )

= ( ac + bd )² + ( ad - bc )²

\(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2=\left(a^2c^2+2abcd+b^2d^2\right)\)\(+\left(a^2d^2-2abcd+b^2c^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\left(đpcm\right)\)

oyfngg

em mới hc lớp 7 à

\(a^2+b^2+c^2+d^2\) 

\(=''a+b''=''b+c''=''c+d''\) 

P/s; Ko chắc đâu nhé mới lớp 5 thôi

Dấu ngoặc kép thay cho dấu ngoặc đơn nha

\(\left(a+b+c\right)^2+a^2+b^2+c^2\)

\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)

\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)

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

2(a-b)(c-b)+2(b-a)(c-a)+2(b-c)(a-c)

=2a^2+2b^2+2c^2-2bc-2ab-2ac

=a^2-2ac+c^2+a^2-2ab+b^2+b^2-2bc+c^2

=(a-c)^2+(a-b)^2+(b-c)^2

a, 1-2x+x^2 = x^2 - 2x.1 + 1^2= (x-1)^2

b, 4y+4+y^2 = y^2 + 2y.2+ 2^2 = (y+2)^2

c, 1/16+1/2x+x^2 = x^2 + 2.x.\(\frac{1}{4}\)+ (1/4)^2 = (x+1/4)^2

d, 36x^2+12xy+y^2 = (6x)^2 + 2.6x.y + y^2 = (6x+y)^2

a) \(1-2x+x^2=\left(1-x\right)^2=\left(x-1\right)^2\)

b) \(4y+4+y^2=y^2+4y+4=\left(y+2\right)^2\)

c) \(\frac{1}{16}+\frac{1}{2}x+x^2=\left(x+\frac{1}{4}\right)^2\)

d) \(36x^2+12xy+y^2=\left(6x+y\right)^2\)