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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.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
a) \(=x^2+7x-12x-84-2x+14\)
\(=x^2-7x-70\)
b)\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
c) \(=9x\left(x+y\right)-\left(x+y\right)\)
\(=\left(9x-1\right)\left(x+y\right)\)
d)\(=\left(x-y\right)^2-9^2\)
\(=\left(x-y+9\right)\left(x-y-9\right)\)
e)\(=x^2+8x+16-60+15x\)
\(=x^2+23x-44\)
\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
\(a,=x\left(x-2\right)\\ b,=2b\left(x-3y\right)+a\left(x-3y\right)=\left(a+2b\right)\left(x-3y\right)\\ c,=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\\ d,=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\\ đ,=5\left(x-y\right)\left(x+y\right)+3\left(x+y\right)^2=\left(x+y\right)\left(5x-5y+3x+3y\right)\\ =\left(x+y\right)\left(8x-2y\right)=2\left(4x-y\right)\left(x+y\right)\\ e,=3x\left(2xy-3\right)\\ b,=x\left(4x^2-4xy+y^2-4\right)=x\left[\left(2x-y\right)^2-4\right]=x\left(2x-y-2\right)\left(2x-y+2\right)\\ f,=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)
\(a,5x^2y-10xy^2=5xy\left(x-2y\right)\\ b,x^2+2xy+y^2-5x-5y=\left(x+y\right)^2-5\left(x+y\right)=\left(x+y\right)\left(x+y-5\right)\\ c,x^2-6x+8=\left(x^2-2x\right)-\left(4x-8\right)=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\\ d,5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\)
c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)
d) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)
e) \(x^2-y^2+2x+1=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
f) \(x^2-2xy-9+y^2=\left(x^2-2xy+y^2\right)-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
c: \(5x^2+15x+3y+xy\)
\(=5x\left(x+3\right)+y\left(x+3\right)\)
\(=\left(x+3\right)\left(5x+y\right)\)
d: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
e: \(x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1-y\right)\left(x+1+y\right)\)
f: \(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-9\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
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. $x^2-y^2+x-y=(x^2-y^2)+(x-y)=(x-y)(x+y)+(x-y)$
$=(x-y)(x+y+1)$
Bài 1:
c) \(x^2-9x+8\)
\(=x^2-8x-x+8\)
\(=x\left(x-8\right)-\left(x-8\right)\)
\(=\left(x-8\right)\left(x-1\right)\)
d)\(x^2+6x+8\)
\(=x^2+4x+2x+8\)
\(=x.\left(x+4\right)+2.\left(x+4\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
Bài 4:
Áp dụng hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b^2\)
=> \(x^2+y^2+2xy+1=\left(x+y\right)^2+1\)
Ta có: \(\left(x+y\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x+y\right)^2+1\ge1\) hay \(\left(x+y\right)^2+1>0\)(đpcm)
Vậy...