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a )x2+2y2-2xy+2x-4y+2=0
<=>x2-2x(y-1)+y2-2y+1+y2-2y+1=0
<=>x2-2x(y-1)+(y-1)2+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>x-y+1=0 va y-1=0
<=>x=y-1 y=1
<=>x=1-1=0 y=1
a) bt \(=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x+1\right)\left(x-2\right)\)
kl: ...
b) \(=\left(x+2\right)\left(x^2-8x-15\right)=\left(x+2\right)\left(x-5\right)\left(x-3\right)\)
kl:....
a, \(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)
\(=\left(x-8\right)\left(x^2-x-2\right)\)
\(=\left(x-8\right)\left(x^2-2x+x-2\right)\)
\(=\left(x-8\right)\left[x\left(x-2\right)+\left(x-2\right)\right]\)
\(=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
b, \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-6\right)\)
\(=\left(x-5\right)\left(x^2-3x+2x-6\right)\)
\(=\left(x-5\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]\)
\(=\left(x-5\right)\left(x-3\right)\left(x+2\right)\)
Chúc bạn học tốt!!!
a \(=9x^2-6x+1+2012\)
\(=\left(3x-1\right)^2+2012\)
\(=200000^2+2012\)
b: \(=2014^2-2\cdot2014\cdot1014+1014^2\)
\(=\left(2014-1014\right)^2=1000^2=10^6\)
c: \(x^2+3y^2=4xy\)
=>x^2-4xy+3y^2=0
=>(x-y)*(x-3y)=0
=>x=y hoặc x=3y
KHi x=y thì \(C=\dfrac{2x+2013x}{x-2x}=-2015\)
Khi x=3y thì \(C=\dfrac{6y+2013y}{3y-2y}=2019\)
\(9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1>0\Rightarrowđpcm\)
\(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(đpcm\right)\)
\(25x^2-20x+7=25x^2-20x+4+3=\left(5x-2\right)^2+3>0\left(đpcm\right)\)
\(9x^2-6xy+2y^2+1=\left(9x^2+6xy+y^2\right)+y^2+1=\left(3x+y\right)^2+y^2+1>0\left(đpcm\right)\)
\(\Leftrightarrow x^2+y^2\ge xy;x^2+y^2\ge2\sqrt{x^2y^2}=2\left|xy\right|\ge\left|xy\right|\ge xy\Rightarrowđpcm\)
\(ĐKXĐ:x\ne\pm\frac{3}{2};x\ne1;x\ne0\)
\(A=\left(\frac{2+3x}{2-3x}-\frac{36x^2}{9x^2-4}-\frac{2-3x}{2+3x}\right):\frac{x^2-x}{2x^2-3x^3}\)
\(=\left[\frac{\left(2+3x\right)^2}{\left(2+3x\right)\left(2-3x\right)}+\frac{36x^2}{\left(2-3x\right)\left(2+3x\right)}-\frac{\left(2-3x\right)^2}{\left(2-3x\right)\left(2+3x\right)}\right]:\frac{x\left(x-1\right)}{x^2\left(2-3x\right)}\)
\(=\frac{4+12x+9x^2+36x^2-4+12x-9x^2}{\left(2+3x\right)\left(2-3x\right)}\cdot\frac{x\left(2-3x\right)}{x-1}\)
\(=\frac{36x^2+24x}{\left(2+3x\right)\left(2-3x\right)}\cdot\frac{x\left(2-3x\right)}{x-1}\)
\(=\frac{12x\left(3x+2\right)}{2+3x}\cdot\frac{x}{x-1}\)
\(=\frac{12x^2}{x-1}\)
Để A nguyên dương hay \(\frac{12x^2}{x-1}\) nguyên dương
Mà \(12x^2\ge0\Rightarrow x-1>0\Rightarrow x>1\)
Vậy để A nguyên dương thì x là số nguyên dương lớn hơn 1.
a) x3 + 2x2y + xy2– 9x = x(x2 +2xy + y2 – 9)
= x[(x2 + 2xy + y2) – 9]
= x[(x + y)2 – 32]
= x(x + y – 3)(x + y + 3)
b) 2x – 2y – x2 + 2xy – y2 = (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)
c) x4 – 2x2 = x2(x2 – (√2)2) = x2(x - √2)(x + √2).
a) 9x4+22+6x2+y2+2y
= (3x2)2+2.3x2.1+1+y2+2y+1+20
=(3x2+1)2 + (y+1)2+22+42
b)x4+4+4y2+5x2+4xy
=x4+5x2+4+4y2+4xy
=x4+4x2+4+4y2+4xy+x2
=(x2)2+2x22+22+(2y)2+2.2yx+x2
=(x2+2)2+(2y+x)2
c)z2+y2-6z+2y+10
=z2-6z+9+y2+2y+1
=z2-2.z.3+9+(y+1)2
=(z-3)2+(y+1)2
d)x2+4y2+m2+4mn+4xy+4n2
=x2+4xy+4y2+4n2+4mn+m2
=x2+2x2y+(2y)2+(2n)2+2.2nm+m2
=(x+2y)2+(2n+m)2
e)x2+y2-6nx+9n2+4my+4m2
=x2-6nx+9n2+y2+4my+4m2
=x2-2x3n+(3n)2+y2+2y2n+(2m)2
=(x-3n)2+(y+2m)2
f)4x2-4xm+2m2+4mn+4n2
=4n2-4xm+m2+4n2+4mn+m2
=(2n)2-2.2xm+m2+(2n)2+2.2nm+m2
=(2n-m)2+(2n+m)2
g) Ghi thiếu đề,đề đúng :
9x2-12xy+5y2+2y+1
=9x2-12xy+4y2+y2+2y+1
=(3x)2-2.3x2y+(2y)2+(y+1)2
=(3x-2y)2+(y+1)2
a) $9x^2+6xy+y^2$
$=(3x)^2+2.3xy+y^2$
$=(3x+y)^2$
b) $6x-9-x^2$
$=-(x^2-6x+9)$
$=-(x-3)^2$
c) $x^2+4y^2+4xy$
$=x^2+(2y)^2+4xy$
$=(x+2y)^2$
d) $(x-2y)^2-(x+2y)^2$
$=(x-2y-x-2y)(x-2y+x+2y)$
$=-4y.2x=-8xy$
a, \(9x^2+6xy+y^2\)
\(=9x^2+3xy+3xy+y^2\)
\(=3x\left(3x+y\right)+y\left(3x+y\right)\)
\(=\left(3x+y\right)^2\)
b, \(6x-9-x^2\)
\(=-\left(x^2-6x+9\right)=-\left(x^2-3x-3x+9\right)\)
\(=-\left(x-3\right)^2\)
c, \(x^2+4y^2+4xy\)
\(=x^2+2xy+2xy+4y^2\)
\(=x\left(x+2y\right)+2y\left(x+2y\right)\)
\(=\left(x+2y\right)^2\)
d, \(\left(x-2y\right)^2-\left(x+2y\right)^2\)
\(=\left(x-2y-x-2y\right)\left(x-2y+x+2y\right)\)
\(=-8xy\)
Chúc bạn học tốt!!!