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a.
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b.
$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$
c.
$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$
d.
$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$
$=(x+1)(x^2-4x+1)$
e.
$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$
$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
f.
$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$
$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$
g.
$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$
$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$
$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$
$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$
h.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
a) x6 – x4 + 2x3 + 2x2
= x2(x4 – x2 + 2x + 2)
= x2[x2(x2 – 1) + 2(x + 1)]
= x2. [x2.(x -1).(x + 1) + 2(x+ 1)]
= x2 (x+ 1).[x2(x- 1)+ 2]
= x2(x + 1)(x3 – x2 + 2)
= x2(x + 1)[(x3 + 1) – (x2 – 1)]
= x2(x + 1).[(x + 1).(x2 – x + 1) - (x - 1).(x + 1)]
= x2(x + 1)(x + 1)( x2 – x + 1 – x + 1)
= x2(x + 1)2(x2 – 2x + 2).
b) 4x4 + y4 = 4x4 + 4x2y2 + y4 - 4x2y2
= (2x2 + y2)2 - (2xy)2
= (2x2 + y2 + 2xy)(2x2 + y2 - 2xy)
\(=\left(x^6+2x^5+x^4\right)-2\left(x^5+2x^4+x^3\right)+2\left(x^4+2x^3+x^2\right)\)
\(=x^2\left(x^2+x\right)^2-2x\left(x^2+x\right)^2+2\left(x^2+x\right)^2\)
\(=\left(x^2+x\right)^2\left(x^2-2x+2\right)\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
Lời giải:
$2x^2+y^2+2xy-6x-2y=8$
$\Leftrightarrow (x^2+y^2+2xy)+x^2-6x-2y=8$
$\Leftrightarrow (x+y)^2-2(x+y)+x^2-4x=8$
$\Leftrightarrow (x+y)^2-2(x+y)+1+(x^2-4x+4)=13$
$\Leftrightarrow (x+y-1)^2+(x-2)^2=13$
$\Rightarrow (x-2)^2=13-(x+y-1)^2\leq 13$
Mà $(x-2)^2$ là scp với mọi $x$ nguyên nên $(x-2)^2\in\left\{0; 1; 4; 9\right\}$
Nếu $(x-2)^2=0\Rightarrow (x+y-1)^2=13-(x-2)^2=13$ (không là scp - loại)
Nếu $(x-2)^2=1\Rightarrow (x+y-1)^2=12$ (không là scp - loại)
Nếu $(x-2)^2=4\Rightarrow (x+y-1)^2=9$
$\Rightarrow x-2=\pm 2$ và $x+y-1=\pm 3$
TH1: $x-2=2; x+y-1=3\Rightarrow x=4; y=0$
TH2: $x-2=2; x+y-1=-3\Rightarrow x=4; y=-6$
TH3: $x-2=-2; x+y-1=3\Rightarrow x=0; y=4$
TH4: $x-2=-2; x+y-1=-3\Rightarrow x=0; y=-2$
Nếu $(x-2)^=9\Rightarrow (x+y-1)^2=4$ (bạn cũng làm tương tự trên)
P = x6 + y6 = (x2 + y2)(x4 - x2 y2 + y4)
= (x2 + y2)2 - 3x2 y2 \(\ge1-3×\frac{\left(x^2+y^2\right)^2}{4}=1-\frac{3}{4}=\frac{1}{4}\)
Đạt được khi x2 = y2 = \(\frac{1}{2}\)
1: \(=\dfrac{-\left[\left(x+5\right)^2-9\right]}{\left(x+2\right)^2}=\dfrac{-\left(x+5-3\right)\left(x+5+3\right)}{\left(x+2\right)^2}\)
\(=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+8\right)}{x+2}\)
2: \(=\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\dfrac{2x}{x+4}\)
3: \(=\dfrac{5x\left(x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{5x}{x^2-1}\)
4: \(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
5: \(=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}=\dfrac{2a\left(a-b\right)}{\left(c+d\right)\left(a-b\right)}=\dfrac{2a}{c+d}\)
6: \(=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\cdot\left(-1\right)=\dfrac{-x}{x+y}\)
7: \(=\dfrac{2\left(1-a\right)}{-\left(1-a^3\right)}=\dfrac{-2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=-\dfrac{2}{1+a+a^2}\)
8: \(=\dfrac{x^4\left(x^3-1\right)}{\left(x^3-1\right)\left(x^3+1\right)}=\dfrac{x^4}{x^3+1}\)
9: \(=\dfrac{\left(x+2-x+2\right)\left(x+2+x-2\right)}{16x}=\dfrac{4\cdot2x}{16x}=\dfrac{1}{2}\)
10: \(=\dfrac{0.5\left(49x^2-y^2\right)}{0.5x\left(7x-y\right)}=\dfrac{1}{x}\cdot\dfrac{\left(7x-y\right)\left(7x+y\right)}{7x-y}\)
\(=\dfrac{7x+y}{x}\)
x=1,y=2
x=0,y=0