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

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

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

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

\(=2b.2a=4ab\) 

a: \(\left(x+y\right)^2-2xy\)

\(=x^2+2xy+y^2-2xy\)

\(=x^2+y^2\)

b: \(\left(a+b\right)^2-\left(a-b\right)\left(a+b\right)\)

\(=\left(a+b\right)\left(a+b-a+b\right)\)

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

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

\(=\left(a+b-a+b\right)\left(a+b+a-b\right)\)

\(=4ab\)

a) Biến đổi VT . Mẫu chung là ( a + 2b )( a - 2b )

\(VT=\frac{a+2b-6b-2\left(a-2b\right)}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 1 )

Biến đổi VP 

\(-\frac{1}{2a}\left(\frac{a^2+4b^2}{a^2-4b^2}+1\right)=-\frac{1}{2a}\cdot\frac{a^2+4b^2+a^2-4b^2}{a^2-4b^2}\)

\(=-\frac{1}{2a}\cdot\frac{2a^2}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 2 )

Từ ( 1 ) và ( 2 ) => VT = VP ( đpcm )

b) \(a^3+b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)^3\)

<=> \(b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)^3=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)-a^3\)( * )

Biến đổi VT của ( * ) ta có :

\(VT=\left[b+\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right]\left[b^2-\frac{b^2\left(2a^3+b^3\right)}{a^3-b^3}+\frac{b^2\left(2a^3+b^3\right)^2}{\left(a^3-b^3\right)^2}\right]\)

\(=\frac{3a^3b}{a^3-b^3}\cdot\frac{3a^6b^2+3a^3b^5+3b^8}{\left(a^3-b^3\right)^2}\)

\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 1 )

\(VP=\left[\frac{a\left(a^3+2b^3\right)}{a^3-b^3}-a\right]\left[\frac{a^2\left(a^3+2b^3\right)^2}{\left(a^3-b^3\right)^2}+\frac{a^2\left(a^3+2b^3\right)}{a^3-b^3}+a^2\right]\)

\(=\frac{3ab^3}{a^3-b^3}\cdot\frac{3a^8+3a^5b^3+3a^2b^6}{\left(a^3-b^3\right)^2}\)

\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 2 )

Từ ( 1 ) và ( 2 ) => VT = VP => ( * ) đúng 

=> Hằng đẳng thức đúng 

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

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

\(VP=VT\Rightarrowđpcm\)

b) Ta có :

\(VT=\left(4x-3y+2\right)-\left(3x-4y+2\right)\)

\(=4x-3y+2-3x+4y-2\)

\(=\left(4x-3x\right)-\left(3y-4y\right)+\left(2-2\right)\)

\(=x+y\)

\(VP=\left(2x+2y\right)-\left(x+y\right)=2x+2y-x-y\)

\(=\left(2x-x\right)+\left(2y-y\right)\)

\(=x+y\)

\(\Rightarrow VT=VP\)

\(\Rightarrow\)đpcm

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

VT : (a + b + c)² + a² + b² + c²

= a² + b² + c² + 2ab +2bc + 2ac + a² + b² + c²

= ( a² + 2ab + b² ) + (b² + 2bc + c²) + ( a² + 2ac + c²)

= (a + b)² + (b + c)² + (a + c)² = VP

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

Ta có : \(VT=\frac{\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2a-b\right)^2}{9}\)

\(=\frac{4a^2+4b^2+8ab+c^2-4ac-4ab+4b^2+4c^2+8bc+a^2-4ba-4bc+4c^2+4a^2+8ac+b^2-4bc-4ab}{9}\)\(=\frac{9\left(a^2+b^2+c^2\right)}{9}=a^2+b^2+c^2=VP\)

Vậy ta có đẳng thức: 

\(\left(\frac{2a+2b-c}{3}\right)^2+\left(\frac{2b+2c-a}{3}\right)^2+\left(\frac{2c+2a-b}{3}\right)^2=a^2+b^2+c^2\)

Bất đẳng thức cần chứng minh tương đương:

\(a^{10}b^2+b^{10}a^2\ge a^8b^4+b^8a^4\)

\(\Leftrightarrow a^8+b^8\ge a^6b^2+b^6a^2\) (Do \(a^2b^2\ge0\))

\(\Leftrightarrow\left(a^6-b^6\right)\left(a^2-b^2\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2\left(a^4+a^2b^2+b^4\right)\ge0\) (luôn đúng).

Vậy ta có đpcm.

\(a^8+b^8-a^6b^2-a^2b^6=\left(a^8-a^6b^2\right)+\left(b^8-a^2b^6\right)=a^6\left(a^2-b^2\right)+b^6\left(b^2-a^2\right)=\left(a^6-b^6\right)\left(a^2-b^2\right)\) nên suy ra được như vậy Quỳnh Anh