\(27y^3-9y^2+y-\frac{1}{27}\)

\(=\left(3y\right)^3-3.\left(3y\right)^2.\frac{1}{3}+3.3y.\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^3\)

\(=\left(3y-\frac{1}{3}\right)^3\)

hk

tốt

2xy + x² + y² = ( x + y )² = ( x+y ) ( x + y )

( 25z -15n )² = ( 25z - 15n ) ( 25z -15n )

\(\left(A+B\right)^2=A^2+2AB+B^2\)

\(\left(A-B\right)^2=A^2-2AB+B^2\)

\(2xy+x^2+y^2=x^2+2.x.y+y^2=\left(x+y\right)^2\)\(\left(25z-15n\right)^2=\left(25n\right)^2-2.25z.15n+\left(15n\right)^2=625z^2-375zn+225n^2\)

Phân tích đa thức thành nhân tử mà bạn , Bạn kia sai rồi !

Đọc lại định nghĩa nhé bạn!

dấu ^ là mũ nha mn

\(x^2-2=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)

\(27x^3-a^3b^3\)

\(=\left(3x\right)^3-\left(ab\right)^3\)

\(=\left(3x-ab\right)\left[\left(3x\right)^2+3x\cdot ab+\left(ab\right)^2\right]\)

\(=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)

27x³ - a³b³

= (3x)³ - (ab)³

= (3x - ab)(9x² + 3xab + a²b²)

Mở sách ra, SGK ghi to thế còn j;-;

\(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)

x³+64=x³+4³=(x+4)(x²-4x+4²)=(x+4)(x²-4x+16)

\(10x-25-x^2=-\left(x^2-10x+25\right)\)

\(=-\left(x^2-2.x.5+5^2\right)=-\left(x-5\right)^2\)

10x - 25 - x²

= x²- 10x - 25

= - ( x² +10x +25)

= -(x² + 2.x.5+5² )

= - (x+5 )²