Question

 Chứng minh: \((ac + bd)^2 + (ad – bc)^2 = (a^2 + b^2)(c^2 + d^2)\) 

Answer 1

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

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

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

Answer 2

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

                                    \( = (ac)^2 +(bd)^2 + (ad)^2 +(bc)^2 + 2abcd – 2abcd\)

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

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

➤ \((ac + bd)^2 + (ad – bc)^2 = (a^2 + b^2)(c^2 + d^2)\)

Answer 3

Bài làm của mình làm như này có dễ hiểu không?

Answer 4

\(\left(a^2+b^2\right)\left(c^2+d^2\right)\\ \Leftrightarrow\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2\\ \Leftrightarrow[\left(ac\right)^2+2abcd+\left(bd\right)^2]+[\left(ad\right)^2-2abcd+\left(bc\right)^2]+2abcd-2abcd\\ \Leftrightarrow\left(ac+bd\right)^2+\left(ad-bc\right)^2\\ \Rightarrowđpcm\)