d: \(9x^2-25=\left(3x-5\right)\left(3x+5\right)\)

e: \(16x^2-4y^2=4\left(4x^2-y^2\right)=4\left(2x-y\right)\left(2x+y\right)\)

f: \(\left(x+\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)

\(d,9x^2-25=\left(3x-5\right)\left(3x+5\right)\\ e,16x^2-4y^2=\left(4x-2y\right)\left(4x+2y\right)=4\left(2x-y\right)\left(2x+y\right)\)

d: \(9x^2-25=\left(3x-5\right)\left(3x+5\right)\)

e: \(16x^2-4y^2=4\left(4x^2-y^2\right)=4\left(2x-y\right)\left(2x+y\right)\)

đề là gì?

\(=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\)

x²+5x+4

(x²+5x)+4

x(x+5)+4

a: \(=-3xy\left(1-2x\right)\)

c: \(=\left(2x-y-5\right)\left(2x-y+5\right)\)

b) (a + b)3 – (a – b)3

(Xuất hiện hằng đẳng thức (7))

= [(a + b) – (a – b)][(a + b)2 + (a + b).(a – b) + (a – b)2]

= (a + b – a + b)(a2 + 2ab + b2 + a2 – b2+ a2 – 2ab + b2)

= 2b.(3a2+ b2)

c) (a + b)3 + (a – b)3

(Xuất hiện hằng đẳng thức (6))

= [(a + b) + (a – b)][(a + b)2 – (a + b)(a –b) + (a – b)2]

= [(a + b) + (a – b)][(a2 + 2ab + b2) – (a2 – b2) + (a2 – 2ab + b2)]

= (a + b + a – b)(a2 + 2ab + b2 – a2 + b2 + a2 – 2ab + b2)

= 2a.(a2 + 3b2)

d) 8x3 + 12x2y + 6xy2 + y3

= (2x)3 + 3.(2x)2.y + 3.2x.y2 + y3

(Xuất hiện hằng đẳng thức (4))

= (2x + y)3

e) –x3 + 9x2 – 27x + 27

= (–x)3 + 3.(–x)2.3 + 3.(–x).32 + 33

(Xuất hiện Hằng đẳng thức (4))

= (–x + 3)3

= (3 – x)3

a) 3x – 6y

= 3.x – 3.2y

(Xuất hiện nhân tử chung là 3)

= 3(x – 2y)

c) 14x2y – 21xy2 + 28x2y2

= 7xy.2x – 7xy.3y + 7xy.4xy

(Xuất hiện nhân tử chung 7xy)

= 7xy(2x – 3y + 4xy)

e) 10x(x – y) – 8y(y – x)

(Nhận thấy x – y = –(y – x) nên ta đổi y – x về x – y)

= 10x(x – y) – 8y[–(x – y)]

= 10x(x – y) + 8y(x – y)

= 2(x – y).5x + 2(x – y).4y

(Xuất hiện nhân tử chung 2(x – y))

= 2(x – y)(5x + 4y)

a) `16-24x+9x^2-25y^2`
`=[4^2-2.4.3x+(3x)^2]-25y^2`
`=(4-3x)^2-(5y)^2`
`=(4-3x+5y)(4-3x-5y)`
b) `x^2-5xy-36y^2`
`=x^2-9xy+4xy-36y^2`
`=x(x-9y)+4y(x-9y)`
`=(x-9y)(x+4y)`

Lời giải:
$E=x(y^2+z^2+2yz)+y(z^2+x^2+2zx)+z(x^2+y^2+2xy)-4xyz$

$=x^2y+xy^2+yz^2+y^2z+z^2x+zx^2+2xyz$

$=xy(x+y)+yz(y+z)+zx(x+z)+2xyz$

$=xy(x+y+z)+yz(y+z+x)+zx(z+x)$

$=(x+y+z)(xy+yz)+zx(z+x)=(x+y+z).y(x+z)+xz(x+z)$

$=(x+z)[y(x+y+z)+xz]$
$=(x+z)[y(x+y)+(yz+xz)]$

$=(x+z)[y(x+y)+z(y+x)]=(x+z)(x+y)(y+z)$

d: \(\left(z-3y\right)^3-\left(z+3y\right)^3\)

\(=\left[\left(z-3y\right)-\left(z+3y\right)\right]\left[\left(z-3y\right)^2+\left(z-3y\right)\left(z+3y\right)+\left(z+3y\right)^2\right]\)

\(=\left(z-3y-z-3y\right)\left(z^2-6zy+9y^2+z^2-9y^2+z^2+6zy+9y^2\right)\)

\(=\left(-6y\right)\cdot\left(3z^2+9y^2\right)\)

\(=-18y\cdot\left(z^2+3y^2\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)