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\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\\ b,=\left(x+y\right)\left(x-5\right)\\ c,=5x^2\left(x-y\right)-10x\left(x-y\right)=5x\left(x-2y\right)\left(x-y\right)\\ d,=x^2-2xy=x\left(x-2y\right)\\ e,=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-y^2-3x+3y\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x^2-y^2\right)\)
\(=2\left(x+y\right)-\left(x-y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x+y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=x^2+y^2+2xy-16\)
\(=\left(x+y\right)^2-16\)
\(=\left(x+y+4\right)\left(x+y-4\right)\)
a) \(x^2-y^2-3x+3y\)
\(=\left(ax+y\right)\left(ax-y\right)-3.\left(x-y\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x+y\right)\left(x-y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=\left(x+y\right)\left(x-y\right)+2xy-16\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (5x - 3y)(x - 3y)(x + 3y).
b) 3(a – b + c) ( x + 6 y ) 2 .
c) (x-y-2m + n)(x-y + 2m-n)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(A=3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)b) \(A=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
c) \(A=x^2+y^2+2xy+yz+zx=\left(x+y\right)^2+z\left(x+y\right)=\left(x+y\right)\left(x+y+z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\frac{x}{y}$ không phải đơn thức bạn nhé.
a. $x^2-2x+1=(x-1)^2$
b. $x^2+2xy-25+y^2=(x^2+2xy+y^2)-25=(x+y)^2-5^2=(x+y-5)(x+y+5)$
c. $5x^2-10xy=5x(x-2y)$
d. $x^2-y^2+x-y=(x^2-y^2)+(x-y)=(x-y)(x+y)+(x-y)$
$=(x-y)(x+y+1)$
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
lên mạng mà kiếm
a, Ta có \(a.\left(a-2\right)-\left(2-a\right).b-\left(a-2\right).2=5\)
\(\Rightarrow\left(a-2\right).\left(a-2\right)-\left(2-a\right).b=5\)
\(\Rightarrow\left(a-2\right).\left[-\left(2-a\right)\right]-\left(2-a\right).b=5\)
\(\Rightarrow\left[-\left(a-2\right)\right].\left(2-a\right)-\left(2-a\right).b=5\)
\(\Rightarrow\left(2-a\right).\left\{\left[-\left(a-2\right)\right]-b\right\}=5\)
Dễ rùi nha , bài 2 chịu