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

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

\(a+b=-c\Rightarrow a^2+b^2+2ab=c^2\)

\(\Rightarrow a^2+b^2=c^2-2ab\)

\(\Rightarrow a^4+b^4+2a^2b^2=c^4+4a^2b^2-4abc^2\)

\(\Rightarrow a^4+b^4=c^4+2a^2b^2-4abc^2\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)=2\left(c^4+2a^2b^2-4abc^2+c^4\right)=4\left(c^4+a^2b^2-2abc^2\right)\)

\(=4\left(c^2-ab\right)^2=\left(2c^2-2ab\right)^2\)

Ta có: (a+b+c)^2 + a^2 + b^2 + c^2

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

= (a^2 +2ab+ b^2) + (b^2 +2bc+ c^2) +(c^2 +2ac+ a^2 )

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

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

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

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

( a² + b² )( c² + d² )

= a²c² + a²d² + b²c² + b²d²

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

= ( ac + bd )² + ( ad - bc )²