\(\left(3+xy^2\right)^2=9+6xy^2+x^2y^4\)

hẳng đằng thức số 1 bạn ạ 

Viết biểu thức sau dưới dạng bình phương của 1 tổng :

\(\left(3+xy^2\right)^2\)

\(=3^2+2.3.\left(xy\right)^2+\left(xy^2\right)^2\)

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

2)
a)
\(=x^3+125\)
\(\)b)\(=1-x^3\)
c)
=\(y^3+27t^3\)
 

Câu 21: 

\(x^2+4x+4=\left(x+2\right)^2\)

Câu 22:

\(x^2-8x+16=\left(x-4\right)^2\)

\(\left(x^2+2x-1\right)^2\)

\(=\left(x^2+2x\right)^2-2\left(x^2+2x\right)+1\)

\(=x^4+4x^3-2x^2+4x^2+4x+1\)

\(=x^4+4x^3+2x^2+4x+1\)

Khó hiểu quá

( 1/2 + x )² = ( 1/2 )² + 2.1/2.x + x² = x² + x + 1/4

( 2x + 1 )² = ( 2x )² + 2.2x.1 + 1² = 4x² + 4x + 1

(x+1)(x^2 - x + 1)

=x^3 - x^2 + x + x^2 -x +1

= x^3 +1

\(\left(x+1\right)\left(x^2-x+1\right)\)

\(=x^3+1\)

Hok tốt !

\(56.64=\left(60-4\right)\left(60+4\right)=60^2-4^2=3600-16=3584\)