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Bài làm:
a) \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(\left(x-y\right)\left(x-y-z\right)\)
a/ \(x^2-2xy+y^2-zx+yz.\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c/ \(x^2-y^2-2x-2y.\)
\(=x^2-2x+1-y^2-2y-1\)
\(=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-1+y+1\right)\left(x-1-y-1\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
\(1.2x^2-2y^2-6x-6y=2\left(x^2-y^2\right)-6\left(x+y\right)\)
\(=2\left(x+y\right)\left(x-y\right)-6\left(x+y\right)=\left(x+y\right)[2\left(x-y\right)-6]=\left(x+y\right)\left(2x-2y-6\right)\)
\(2.x^4+x^3-x^2-x=\left(x^4+x^3\right)-\left(x^2+x\right)\)
\(=x^3\left(x+1\right)-x\left(x+1\right)=\left(x+1\right)\left(x^3-x\right)\)
\(3.a^3+a^2b-a^2c-abc\)( mình trả lời ở câu hỏi của bạn rồi)
\(4.x^5-x^3+x^2-1=\left(x^5-x^3\right)+\left(x^2-1\right)\)
\(=x^3\left(x^2-1\right)+\left(x^2-1\right)=\left(x^2-1\right)\left(x^3+1\right)\)
\(5.x+y\left(x-1\right)-1\) ( mình trả lời ở câu hỏi của bạn rồi)
câu 6 và 7 cũng vậy
a) x3+y3+2x2-2xy+2y2=(x3+y3)+(2x2-2xy+2y2)=(x+y)(x2-xy+y2)+2(x2-xy+y2)=(x+y+2)(x2-xy+y2)
b) tương tự câu a
c) (a2+b2-5)2-4(ab+2)2=(a2+b2-5)2-22(ab+2)2
= (a2+b2-5)2-(2ab+4)2=(a2-2ab+b2-5-4)(a2+2ab+b2-5+4)
=( (a-b)2 -9 )( (a+b)2 -1 ) = (a-b-3)(a-b+3)(a+b-1)(a+b+1)
a)
\(=x^2\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
b)
\(=a\left(a-b\right)+a-b\)
\(=\left(a+1\right)\left(a-b\right)\)
c)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left(x+1-y\right)\left(x+1+y\right)\)
d)
\(=x^3\left(x-2\right)+10x\left(x-2\right)\)
\(=x\left(x^2+10\right)\left(x-2\right)\)
e)
\(=x\left(x^2+2x+1\right)\)
\(=x\left(x+1\right)^2\)
f)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right)\left(x+y\right)\)
a,2x3+3x2+2x+3
=(2x3+2x)+(3x2+3)
=2x(x2+1)+3(x2+1)
=(x2+1)(2x+3)
b,a2-ab+a-b
=(a2-ab)+(a-b)
=a(a-b)+(a-b)
=(a-b)(a+1)
c,2x2+4x+2-2y2
=2(x2+2x+1-y2)
=2[(x2+2x+1)-y2 ]
=2[(x+1)2-y2 ]
=2(x+1-y)(x+1+y)
d,x4-2x3+10x2-20x
=(x4-2x3)+(10x2-20x)
=x3(x-2)+10x(x-2)
=(x-2)(x3+10x)
=(x-2)[x(x2+10)]
e,x3+2x2+x
=x(x2+2x+1)
=x(x+1)2
f,xy+y2-x-y
=(xy+y2)-(x-y)
=y(x+y)-(x+y)
=(x+y)(y-1)
a) \(=x^2+2xy+y^2-x^2+y^2=2xy+2y^2=2y\left(x+y\right)\)
b) \(=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
c) \(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
e) \(=\left(x-3\right)\left(x^2+3x+9\right)-2x\left(x-3\right)=\left(x-3\right)\left(x^2+x+9\right)\)
f) \(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)=\left(x+5\right)\left(x^2-6x+25\right)\)
a) \(\left(x+y\right)^2-\left(x^2-y^2\right)\)
\(=x^2+2xy+y^2-x^2+y^2\)
\(=2y^2+2xy\)
\(=2y\left(x+y\right)\)
c) \(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-x^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)
\(=\left[2x\left(y+1\right)+\left(y+1\right)\right]\left[2x\left(y-1\right)-\left(y-1\right)\right]\)
\(=\left(2x+1\right)\left(y+1\right)\left(2x-1\right)\left(y-1\right)\)
\(=\left(4x^2-1\right)\left(y^2-1\right)\)
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/ \(\left(a^2+2ab+b^2\right)+\left(a+b\right)\)
= \(\left(a+b\right)^2+\left(a+b\right)\)
= \(\left(a+b\right)\left(a+b+1\right)\)
Lời giải:
a)
$x^3+y^3+2x^2-2xy+2y^2=(x^3+y^3)+2(x^2-xy+y^2)$
$=(x+y)(x^2-xy+y^2)+2(x^2-xy+y^2)=(x^2-xy+y^2)(x+y+2)$
b)
$a^4+ab^3-a^3b-b^4=(a^4-a^3b)+(ab^3-b^4)$
$=a^3(a-b)+b^3(a-b)=(a-b)(a^3+b^3)=(a-b)(a+b)(a^2-ab+b^2)$
c)
\(a^3-b^3+3a^2+3ab+3b^2=(a^3-b^3)+3(a^2+ab+b^2)\)
\(=(a-b)(a^2+ab+b^2)+3(a^2+ab+b^2)=(a^2+ab+b^2)(a-b+3)\)
d)
\(x^4+x^3y-xy^3-y^4=x^3(x+y)-y^3(x+y)=(x+y)(x^3-y^3)=(x+y)(x-y)(x^2+xy+y^2)\)