Question

7 Chứng minh các đẳng thức sau

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

c) \(a^6+b^6=\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]\)

d) \(a^6-b^6=\left(a^2-b^2\right)\left[\left(a^2+b^2\right)^2-a^2b^2\right]\)

Answer 1

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

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

\(\Rightarrowđpcm\)

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

\(VP=a^4+b^4+2a^2b^2-2a^2b^2=a^4+b^4=VT\)\(\Rightarrowđpcm\)

c) ​\(a^6+b^6=\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]\)

\(VP=\left(a^2+b^2\right)\left(a^4-a^2b^2+b^4\right)=a^6+b^6\)

\(VP=VT\Rightarrowđpcm\)

d)\(a^6-b^6=\left(a^2-b^2\right)[\left(a^2+b^2\right)^2-a^2b^2]\)

\(VP=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)=a^6-b^6=VT\)

\(VP=VT\Rightarrowđpcm\)