x⁴ + y⁴ +(x+y)⁴ = x⁴ + y⁴ + x⁴ + 4x³y + 6x²y² +4xy³ + y⁴ = 2x⁴ +2y⁴ +4x²y²+4x³y+4xy³+2x²y²

= 2(x⁴ +y⁴ +2x²y²)+4xy(x²+y²) + 2x²y²= 2(x² + y²)² + 4xy(x² + y²) +2x²y²

=2((x² +y²) +2xy(x²+ y²) +x²y²) = 2(x² + y² + xy)² \(\Rightarrow\)  đpcm

BTBTVP, ta có:

\(2\left(x^2+xy+y^2\right)^2\)

= \(2x^4+2x^2y^2+2y^4\)

= \(x^4+x^4+2x^2y^2+y^4+y^4\)

= \(x^4+y^4+\left(x^2+y^2\right)^2\)

=\(x^4+y^4+\left[\left(x+y\right)^2\right]^2\)

= \(x^4+y^4+\left(x+y\right)^4\)

x⁴ + y⁴ +(x+y)⁴ = x⁴ + y⁴ + x⁴ + 4x³y + 6x²y² +4xy³ + y⁴ = 2x⁴ +2y⁴ +4x²y²+4x³y+4xy³+2x²y²

= 2(x⁴ +y⁴ +2x²y²)+4xy(x²+y²) + 2x²y²= 2(x² + y²)² + 4xy(x² + y²) +2x²y²

=2((x² +y²) +2xy(x²+ y²) +x²y²) = 2(x² + y² + xy)² \(\Rightarrow\)  đpcm

a) (x-a)^4-(x+a)^4

=[(x-a)^2]^2-[(x+a)^2]^2

=[(x-a)^2-(x+a)^2][(x-a)^2+(x+a)^2]

=[(x-a-x-a)(x-a+x+a)][(x-a)^2+(x+a)^2]

=(-2a.2x)(x^2-2xa+a^2+x^2+2xa+a^2)

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

=-4ax(2x^2+2a^2)

=-4ax.2(x^2+a^2)

a/ (x-a)^4 - (x+a)^4

    =((x-a)^2)^2 - ((x+a)^2)^2

    =(x^2 - 2xa + a^2)^2 - (x^2 +2xa+a^2)^2

    =(x^2-2xa+a^2-x^2-2xa-a^2)(x^2-2xa+a^2+x^2+2xa+a^2)

    =-4xa(2x^2+2a^2)

b/ x^4 –y^2(2x-y)^2

    =(x^2)^2-(y(2x-y)^2

    =(x^2)^2-(2xy-y^2)^2

    =(x^2-2xy+y^2)(x^2+2xy+y^2)

    =(x-y)^2 (x+y)^2

c/(xy+4)^2- 4(x+y)^2

    =(xy+4)^2- (2x+2y)^2

    =(xy+y-2x-2y)(xy+y+2x+2y)

    =(xy-y+2x)(xy+3y+2x)

sử dụng tam giác  Pascal

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

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

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

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

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

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

Tương tự mấy câu kia

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

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

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

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

c: \(x^4+y^4-2\left(x^2+xy+y^2\right)^2\)

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

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

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

=>\(x^4+y^4+\left(x+y\right)^4=2\left(x^2+xy+y^2\right)^2\)

kic cho mik

a, \(\left(x+1\right)^2-25=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

b, \(\left(xy+4\right)^2-4\left(x+y\right)^2=\left(xy+4\right)^2-\left(2x+2y\right)^2=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)

c, xem lại đề nhé 

Ta có: \(x^2\cdot\left(x^4+25\right)\cdot\left(x^2-5\right)\cdot\left(x^2+5\right)\cdot\left(x-y\right)\left(x^2+xy+y^2\right)\cdot\left(x^3+y^3\right)\)

\(=x^2\cdot\left(x^4+25\right)\left(x^4-25\right)\cdot\left(x^3-y^3\right)\left(x^3+y^3\right)\)

\(=x^2\cdot\left(x^8-625\right)\cdot\left(x^6-y^6\right)\)