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18 tháng 7 2021

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

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

c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)

d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)

18 tháng 7 2021

a) Ta có: \(x^2-2x-4y^2-4y\)

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

\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

b) Ta có: \(x^3+2x^2+2x+1\)

\(=\left(x^3+1\right)+2x\left(x+1\right)\)

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

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

AH
Akai Haruma
Giáo viên
25 tháng 10 2021

a. 

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

b.

$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$

c.

$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$

d.

$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$

$=(x+1)(x^2-4x+1)$

AH
Akai Haruma
Giáo viên
25 tháng 10 2021

e.

$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$

$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$

f.

$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$

$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$

g.

$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$

$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$

$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$

$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$

h.

$x^6+2x^5+x^4-2x^3-2x^2+1$

$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$

$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$

6 tháng 10 2025

a: \(x^4+2x^3+3x^2+2x+1\)

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

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

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

b: \(x^4-4x^3+2x^2+4x+1\)

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

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

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

c: \(x^4+x^3+2x^2+2x+4\)

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

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

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

24 tháng 8 2021

a: Ta có: \(x^2-4y^2-2x-4y\)

\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

c: Ta có: \(x^3+2x^2y-x-2y\)

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

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

d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)

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

\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)

\(=\left(x-y\right)\left(x+5y\right)\)

24 tháng 8 2021

e: Ta có: \(x^3-4x^2-9x+36\)

\(=x^2\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

f: Ta có: \(x^2-y^2-2x-2y\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-2\right)\)

18 tháng 7 2021

e) Ta có: \(x^4-2x^3+2x-1\)

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

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

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

\(=\left(x+1\right)\cdot\left(x-1\right)^3\)

h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)

\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)

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

\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)

\(=\left(x-y\right)\left(x+5y\right)\)

18 tháng 7 2021

a) Ta có: \(x^2-y^2-2x-2y\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-2\right)\)

b) Ta có: \(x^2\left(x+2y\right)-x-2y\)

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

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

1 tháng 10 2021

`a)x^3-8x^2+16x`

`=x(x^2-8x+16)`

`=x(x-4)^2`

`b)x^2+4y^2+2x-4y-4xy-24`

`=(x-2y)^2+2(x-2y)-24`

`=(x-2y)^2-4(x-2y)+6(x-2y)-24`

`=(x-2y-4)(x-2y+6)`

`c)x^4+x^3-x^2-2x-2`

`=x^4-2x^2+x^3-2x+x^2-2`

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

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

5 tháng 9 2021

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

 

 

5 tháng 9 2021

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

2:

a: \(3xy^2-3x^3-6xy+3x\)

\(=3x\cdot\left(y^2-2y+1-x^2\right)\)

\(=3x\left\lbrack\left(y-1\right)^2-x^2\right\rbrack\)

=3x(y-1-x)(y-1+x)

b: \(3x^2+11x+6\)

\(=3x^2+9x+2x+6\)

=3x(x+3)+2(x+3)

=(x+3)(3x+2)

c: \(-x^3-4xy^2+4x^2y+16x\)

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

\(=x\cdot\left\lbrack4^2-\left(x^2-4xy+4y^2\right)\right\rbrack=x\cdot\left\lbrack4^2-\left(x-2y\right)^2\right\rbrack\)

=x(4-x+2y)(4+x-2y)

d: \(xz-x^2-yz+2xy-y^2\)

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

=\(z\left(x-y\right)-\left(x-y\right)^2\)

=(x-y)(z-x+y)

e: \(4x^2-y^2-6x+3y\)

=(2x-y)(2x+y)-3(2x-y)

=(2x-y)(2x+y-3)

f: \(x^4-x^3-10x^2+2x+4\)

\(=x^4+2x^3-2x^2-3x^3-6x^2+6x-2x^2-4x+4\)

\(=\left(x^2+2x-2\right)\left(x^2-3x-2\right)\)

g: \(\left(x^3-x^2+x\right)\left(121-25y^2-10y\right)-\left(x^3-x^2+x\right)-\left(121-25y^2-10y\right)+1\)

\(=\left(x^3-x^2+x\right)\left(121-25y^2-10y-1\right)-\left(121-25y^2-10y-1\right)\)

\(=\left(x^3-x^2+x-1\right)\left\lbrack121-\left(25y^2+10y+1\right)\right\rbrack\)

\(=\left(x-1\right)\left(x^2+1\right)\left\lbrack121-\left(5y+1\right)^2\right\rbrack\)

=(x-1)(x^2+1)(11-5y-1)(11+5y+1)

=(x-1)(x^2+1)(10-5y)(12+5y)

=5(2-y)(x-1)(x^2+1)(5y+12)


11 tháng 12 2021

b: \(=x\left(x-3\right)\left(x^2+3x+9\right)\)

23 tháng 12 2022

a/ 2x^2 (x – 1) + 4x (1 – x)

= 2x^2(x  – 1) – 4x (x – 1)

= (x – 1)( 2x^2 – 4x)

=2x(x – 1)(x – 2)

 

1 tháng 10 2021

1/(x+2)-(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x

2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)