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\(x^2-xy+x-y.\)
\(=\left(x^2-xy\right)+\left(x-y\right)\)
\(=x.\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-y\right).\left(x+1\right)\)
a, \(8^3yz+12^2yz+6xyz+yz\)
\(=512yz+144yz+6xyz+yz\)
\(=yz\left(512+14+6x+1\right)\)
\(=yz\left(527+6x\right)\)
$---$
b, \(81x^4\left(z^2-y^2\right)-z^2+y^2\)
\(=81x^4\left(z^2-y^2\right)-\left(z^2-y^2\right)\)
\(=\left(z^2-y^2\right)\left(81x^4-1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(9x^2\right)^2-1^2\right]\)
\(=\left(z-y\right)\left(z+y\right)\left(9x^2-1\right)\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(3x\right)^2-1^2\right]\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left(3x-1\right)\left(3x+1\right)\left(9x^2+1\right)\)
$---$
c, \(\dfrac{x^3}{8}-\dfrac{y^3}{27}+\dfrac{x}{2}-\dfrac{y}{3}\)
\(=\left[\left(\dfrac{x}{2}\right)^3-\left(\dfrac{y}{3}\right)^3\right]+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}\right)+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}+1\right)\)
$---$
d, \(x^6+x^4+x^2y^2+y^4-y^6\)
\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^4+x^2y^2+y^4\right)\left(x^2-y^2+1\right)\)
$Toru$
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a) \(x^2+2x-8=x^2+2x+1-9=\left(x+1\right)^2-3^2=\left(x+1-3\right)\left(x+1+3\right)\)
\(=\left(x-2\right)\left(x+4\right)\)
b) \(x^4+x^2+1=x^4+2x^2+1-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
Bạn k cho mình nhé
Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
\(x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)
ukm theo mình thì:
2a^2 + 4ab + 2b^2
= 2 (a^2 + 2ab + b^2)
= 2 (a + b )^2
ta có: 81x^4 + 4
= (3x)^4 + 2^2
= (9x^2)^2 + 2.9x^2.2+2^2- 36x^2
= (9x^2)^2 + 36x^2 + 2^2 -36x^2
= (9x^2+2)^2 - (6x)^2
= (9x^2+2-6x)( 9x^2 +2 +6x)
ta có: 81x^4 + 4
= (3x)^4 + 2^2
= (9x^2)^2 + 2.9x^2.2+2^2- 36x^2
= (9x^2)^2 + 36x^2 + 2^2 -36x^2
= (9x^2+2)^2 - (6x)^2
= (9x^2+2-6x)( 9x^2 +2 +6x)