a) $(3x+5)^2\\=(3x)^2+2.3x.5+5^2\\=9x^2+30x+25$

b) $(6x+\dfrac{1}{3})^2\\=(6x)^2+2.6x.\dfrac{1}{3}+(\dfrac{1}{3})^2\\=36x^2+4x+\dfrac{1}{9}$

c) $(5x-4y)^2\\=(5x)^2-2.5x.4y+(4y)^2\\=25x^2-40xy+16y^2$

d) $(5x-3)(5x+3)\\=(5x)^2-(3)^2\\=25x^2-9$

= x²( x-6)-5(x-6)

=(x²-5)(x-6)

` @Answer`

`2,(7+3x-y)(7-3x+y)`

`=(7+3x)^2 - y^2`

`3, (6/5 - 5x)^2`

`=(6/5)^2 - 2 . 6/5 . 5x+ (5x)^2`

`= 36/25 - 12x + 25x^2`

Công thức :

`(x-y)(x+y) = x^2 - y^2`

`(x+y)^2=x^2+2xy+y^2`

`+,` Đề `1:(2x+y^2)` ngoài ngoặc có gì ạ ?

2)

(7 + 3x - y)(7 - 3x + y)

= [7 + (3x - y)][7 - (3x + y)]

= 7² - (3x - y)²

= 49 - (3x - y)²

= 49 - (9x² - 6xy + y²)

= 49 - 9x² + 6xy - y²

a,(x+2y)³ =x³+3.x².2y+3.x.(2y)²+(2y)³

= x³+6x²y+12xy²+8y³

b, phần b tương tự dấu thay đổi một tí

c, (5x+1)(5x+1)= (5x+1)²

=25x²+10x+1

a)  \(\left(x+2y\right)^3=x^3+6x^2y+12xy^2+8y^3\)

b)  \(\left(2x-1\right)^3=8x^3-12x^2+6x-1\)

c)  \(\left(5x+1\right)\left(5x-1\right)=25x^2-1\)

bạn Đường Quỳnh Giang phần c bị thiếu 10x rồi

(x+3)(2x²-5x+1)

=2x³-5x²+x+6x²-15x+3

=2x³-5x²+6x²+x-15x+3

=2x³+x²-14x+3

Vậy hệ số x là : -14

\(\left(5x+3yz\right)^2=25x^2+30xyz+9y^2z^2\\ \left(2x-3\right)^3=8x^3-36x^2+54x-27\)

\(\left(y^{2x}+3yz\right)^2=y^{4x}+6y^{2x+1}z+9y^2z^2\\ \left(x^2-6z\right)\left(x^2+6z\right)=x^4-36z^2\\ \left(y-5\right)\left(25+2y+y^2+3y\right)=\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)