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a) a2 + b2 + 2ab + 2a + 2b + 1
= (a2 + b2 + 2ab) + (2a + 2b) + 1
= (a + b)2 + 2(a + b) + 1
= (a + b + 1)2
b) a3 - 3a + 3b - b3
= (a3 - b3) - (3a - 3b)
= (a - b)(a2 - ab + b2) - 3(a - b)
= (a - b)(a2 - ab + b2 - 3)
c) x2 + 2x - 15
= (x2 + 2x + 1) - 16
= (x + 1)2 - 16
= (x + 1 - 5)(x + 1 + 5)
= (x - 4)(x + 6)
d) a4 + 6a2b + 9b2 - 1
= (a2 + 3b)2 - 1
= (a2 + 3b - 1)(a2 + 3b + 1)
a)\(2a^2-3ab+b^2\)
=\(a^2+a^2-2ab-ab+b^2\)
=\(\left(a-b\right)^2+a\left(a-b\right)\)
=\(\left(a-b\right)\left(2a-b\right)\)
b)\(x^2-7x-30\)
=\(x^2-10x+3x-30\)
=\(x\left(x-10\right)+3\left(x-10\right)\)
=\(\left(x-10\right)\left(x+3\right)\)
c)\(6a^2-5ab-6b^2\)
=\(6a^2-9ab+4ab-6b^2\)
=\(3a\left(2a-3b\right)+2b\left(2a-3b\right)\)
=\(\left(2a-3b\right)\left(3a+2b\right)\)
d)\(a^4+a^2+1\)
=\(a^4+2a^2-a^2+1\)
=\(\left(a^2+1\right)^2-a^2\)
=\(\left(a^2+1-a\right)\left(a^2+1+a\right)\)
e)\(x^3+6x^2+11x+6\)
=\(x\left(x^2+6x+9+2\right)+6\)
\(=x\left(\left(x+3\right)^2+2\right)+6\)
=\(x\left(x+3\right)^2+2x+6\)
=\(x\left(x+3\right)^2+2\left(x+3\right)\)
=\(\left(x+3\right)\left(x^2+3x+2\right)\)
a) \(3xy^2-12xy+12x\)
\(=3x\left(y-4y+4\right)\)
b) \(3x^3y-6x^2y-3xy^3-6axy^2-3a^2xy+3xy\)
\(=3xy\left(x^2-2x-y^2-2ay-a^2+1\right)\)
\(=3xy\left[\left(x^2-2\cdot x\cdot1+1^2\right)-\left(y^2+2\cdot y\cdot a+a^2\right)\right]\)
\(=3xy\left[\left(x-1\right)^2-\left(y+a\right)^2\right]\)
\(=3xy\left(x-1-y-a\right)\left(x-1+y+a\right)\)
c) \(36-4a^2+20ab-25b^2\)
\(=6^2-\left[\left(2a\right)^2-2\cdot2a\cdot5b+\left(5b\right)^2\right]\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
d) \(5a^3-10a^2b+5ab^2-10a+10b\)
\(=5a\left(a^2-2ab+b^2\right)-10\left(a-b\right)\)
\(=5a\left(a-b\right)^2-10\left(a-b\right)\)
\(=\left(a-b\right)\left[5a\left(a-b\right)-10\right]\)
\(=5\left(a-b\right)\left[a\left(a-b\right)-2\right]\)
\(=5\left(a-b\right)\left(a^2-ab-2\right)\)
a. 3xy2-12xy+12x
= 3x(y2-4y+4)
= 3x(y-2)2
b. 3x3y-6x2y-3xy3-6axy2-3a2xy+3xy
= 3xy( x2-2x-y2-2ay-a2+1)
= 3xy ((x2-2x+1)-(a2-2ay-y2))
=3xy((x-1)2-(a-y)2)
= 3xy((x-1+a-y)(x-1-(a-y))
=3xy(x-1+a-y)(x-1-a+y)
d. =( 5a(a2-2ab+b2))-(10(a+b))
= 5a(a-b)2-10(a-b)
=5a(a-b)(a-b)-10(a-b)
=(a-b)(5a(a-b)-10)
Hình như mik làm sai hết rồi
1)
b) \(\left(x-z\right)^2-y^2+2y-1\)
\(=\left(x^2-2xz+z^2\right)-\left(y-1\right)^2\)
\(=\left(y-z\right)^2-\left(y-1\right)^2\)
\(=\left[\left(x-z\right)+\left(y-1\right)\right]\cdot\left[\left(x-z\right)-\left(y+1\right)\right]\)
\(=\left(x-z+y-1\right)\cdot\left(x-z-y-1\right)\)
4/ a/ Ta có \(x^2-2xy+y^2+a^2=\left(x-y\right)^2+a^2\)
Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\a^2\ge0\end{cases}}\)=> \(\left(x-y\right)^2+a^2\ge0\)
=> \(x^2-2xy+y^2+a^2\ge0\)
Vậy \(x^2-2xy+y^2\)chỉ nhận những giá trị không âm.
b/ Ta có \(x^2+2xy+2y^2+2y+1=\left(x^2+2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x+y\right)^2+\left(y+1\right)^2\)
Mà \(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)=> \(\left(x+y\right)^2+\left(y+1\right)^2\ge0\)
=> \(x^2+2xy+2y^2+2y+1\ge0\)
Vậy \(x^2+2xy+2y^2+2y+1\)chỉ nhận những giá trị không âm.
c/ Ta có \(9b^2-6b+4c^2+1=\left(3b-1\right)^2+4c^2\)
Mà \(\hept{\begin{cases}\left(3b-1\right)^2\ge0\\4c^2\ge0\end{cases}}\)=> \(\left(3b-1\right)^2+4c^2\ge0\)
=> \(9b^2-6b+4c^2+1\ge0\)
Vậy \(9b^2-6b+4c^2+1\)chỉ nhận những giá trị không âm.
d/ Ta có \(x^2+y^2+2x+6y+10=\left(x+1\right)^2+\left(y+3\right)^2\)
Mà \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)=> \(\left(x+1\right)^2+\left(y+3\right)^2\ge0\)
=> \(x^2+y^2+2x+6y+10\ge0\)
Vậy \(x^2+y^2+2x+6y+10\)chỉ nhận những giá trị không âm.
1/
a/ \(x^4-y^4=\left(x^2-y^2\right)\)
b/ \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left[a^2+2ab+b^2-\left(a^2-b^2\right)+\left(a^2-2ab+b^2\right)\right]\)
\(=2b\left(a^2+b^2\right)\)
c/ \(\left(a^2+2ab+b^2\right)+\left(a+b\right)\)
= \(\left(a+b\right)^2+\left(a+b\right)\)
= \(\left(a+b\right)\left(a+b+1\right)\)
\(a,2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)
\(b,4x-4y+x^2\left(y-x\right)=4\left(x-y\right)-x^2\left(x-y\right)=\left(x-y\right)\left(2-x\right)\left(2+x\right)\)
\(c,2xz+y^2-x^2-z^2=y^2-\left(x^2-2xz+z^2\right)=y^2-\left(x-z\right)^2=\left(y-x+z\right)\left(y+x-z\right)\)
\(d,3a^2-3ab+9b-9a=3a\left(a-b\right)-9\left(a-b\right)=\left(3a-9\right)\left(a-b\right)=3\left(a-3\right)\left(a-b\right)\)
\(f,x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
dat \(x^2-2x+2=y\)
ta co pt
\(y^4+20x^2y^2+64x^4\)
\(=\left(8x^2\right)^2+2.8x^2.\frac{10}{8}y^2+\left(\frac{10^{ }}{8^{ }}y^2\right)^2-\frac{36}{64}y^4\)
\(=\left(8x^2+\frac{10}{8}y^2\right)^2-\left(\frac{6}{8}y^2\right)^2\)
\(=\left(8x^2+\frac{y^2}{2}\right)\left(8x^2+2y^2\right)\)
bạn thay y nữa là xong
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+64x^4\)
\(=\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)^2+100x^4-36x^4\)
\(=\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^4\)
\(=\left(x^4-4x^3+18x^2-8x+4\right)^2-\left(6x^2\right)^2\)
\(=\left(x^4-4x^3+24x^2-8x+4\right)\left(x^4-4x^3+12x^2-8x+4\right)\)
\(\left(x^2-2x+2\right)^4+20x^2\left(x^2-2x+2\right)+64x^4\)
=\(\left[\left(x^2-2x+2\right)^4+2.10x^2\left(x^2-2x+2\right)^2+100x^4\right]\)-100x4+64x2
=\(\left[\left(x^2-2x+2\right)^2+10x^2\right]^2-36x^2\)
=\(\left[\left(x^2-2x+2\right)^2+4x^2\right].\left[\left(x^2-2x+2\right)^2+16x^2\right]\)
a) 4x2 - 20x + 25 - 36y2
= (2x - 5)2 - 36y2
= (2x - 5 - 6y)(2x - 5 + 6y)
b) x3 + x2 - 2x - 8
= (x3 - 8) + (x2 - 2x)
= (x - 2)(x2 + 2x + 4) + x(x - 2)
= (x - 2)(x2 + 2x + 4 + x)
= (x - 2)(x2 + 3x + 4)
d) x4 + 6x3 + 9x2 - 16
= x2(x2 + 6x + 9) - 16
= x2(x + 3)2 - 16
= (x2 + 3x)2 - 16
= (x2 + 3x - 4)(x2 + 3x + 4)
= (x2 + 4x - x - 4)(x2 + 3x + 4)
= [x(x + 4) - (x + 4)](x2 + 3x + 4)
= (x - 1)(x + 4)(x2 + 3x + 4)
A = x2 - 20x - 125
= x2 + 5x - 20x - 125
= x( x + 5 ) - 25( x + 5 )
= ( x + 5 )( x - 25 )
B = 12x2 - 2x - 4
= 12x2 + 6x - 8x - 4
= 6x( x + 2 ) - 4( x + 2 )
= ( x + 2 )( 6x - 4 )
C = 3a2 - 5ab - 12b2
= 3a2 - 9ab + 4ab - 12b2
= 3a( a - 3b ) + 4b( a - 3b )
= ( a - 3b )( 3a + 4b )
D = 25ab - 6a2 + 9b2
= 9b2 + 27ab - 2ab - 6a2
= 9b( b + 3a ) - 2a( b + 3a )
= ( b + 3a )( 9b - 2a )