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a) \(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(=\left(xy+1-x+y\right)\left(xy+1+x-y\right)\)
b) \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(x+y-x+y\right)\left[\left(x^2+2xy+y^2\right)+x^2-y^2+\left(x^2-2xy+y^2\right)\right]\)
\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
c) \(3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=3y^2\left(x^4+x^3+x+1\right)\)
d) \(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\)
\(=4\left[\left(x^2-y^2\right)-2\left(x-ay\right)-\left(a^2-1\right)\right]\)
\(=4\left[\left(x^2-y^2\right)-\left(2x-2ay\right)-\left(a^2-1\right)\right]\)
\(=4\left(x^2-y^2-2x+2ay-a^2+1\right)\)
P/s: Ko chắc!
c/
\(=3y^2\left(x^4+x^3+x+1\right)\)
\(=3y^2\left[x^3\left(x+1\right)+x+1\right]\)
\(=3y^2\left(x^3+1\right)\left(x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
d/
\(=\left(4x^2-8x+4\right)-\left(4y^2-8ay+4a^2\right)\)
\(=4\left(x-1\right)^2-4\left(y-a\right)^2\)
\(=4\left[\left(x-1\right)^2-\left(y-a\right)^2\right]\)
\(=4\left(x-1-y+a\right)\left(x-1+y-a\right)\)
a) Ta có: \(\left(3-xy^2\right)^2-\left(2+xy^2\right)^2\)
\(=\left[\left(3-xy^2\right)-\left(2+xy^2\right)\right]\cdot\left[\left(3-xy^2\right)+\left(2+xy^2\right)\right]\)
\(=\left(3-xy^2-2-xy^2\right)\cdot\left(3-xy^2+2+xy^2\right)\)
\(=5\cdot\left(1-2xy^2\right)\)
\(=5-10xy^2\)
b) Ta có: \(9x^2-\left(3x-4\right)^2\)
\(=\left[3x-\left(3x-4\right)\right]\left[3x+\left(3x-4\right)\right]\)
\(=\left(3x-3x+4\right)\cdot\left(3x+3x-4\right)\)
\(=4\cdot\left(6x-4\right)\)
\(=24x-16\)
c) Ta có: \(\left(a-b^2\right)\left(a+b^2\right)\)
\(=a^2-b^4\)
d) Ta có: \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)
\(=\left(a^2+2a\right)^2-9\)
\(=a^4+4a^3+4a^2-9\)
e) Ta có: \(\left(x-y+6\right)\left(x+y-6\right)\)
\(=x^2+xy-6x-yx-y^2+6y+6x+6y-36\)
\(=x^2-y^2+12y-36\)
f) Ta có: \(\left(y+2z-3\right)\left(y-2z-3\right)\)
\(=\left(y-3\right)^2-\left(2z\right)^2\)
\(=y^2-6y+9-4z^2\)
g) Ta có: \(\left(2y-5\right)\left(4y^2+10y+25\right)\)
\(=\left(2y\right)^3-5^3\)
\(=8y^3-125\)
h) Ta có: \(\left(3y+4\right)\left(9y^2-12y+16\right)\)
\(=\left(3y\right)^3+4^3\)
\(=27y^3+64\)
i) Ta có: \(\left(x-3\right)^3+\left(2-x\right)^3\)
\(=\left(x-3\right)^3-\left(x-2\right)^3\)
\(=x^3-9x^2+27x-27-\left(x^3-6x^2+12x-8\right)\)
\(=x^3-9x^2+27x-27-x^3+6x^2-12x+8\)
\(=-3x^2+15x-19\)
j) Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(x+y\right)-\left(x-y\right)\right]\cdot\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\cdot\left(3x^2+y^2\right)\)
\(=6x^2y+2y^3\)
a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)
b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)
c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)
d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2
= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)
e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)
f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)
g) chắc là 3xyz
= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)
h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)
i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy
k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).
a) Ta có: \(4\left(2-x\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+y\left(x-2\right)\)
\(=\left(x-2\right)\left[4\left(x-2\right)+y\right]\)
\(=\left(x-2\right)\left(4x-8+y\right)\)
b) Ta có: \(3a^2x-3a^2y+abx-aby\)
\(=3a^2\left(x-y\right)+ab\left(x-y\right)\)
\(=\left(x-y\right)\left(3a^2+ab\right)\)
\(=a\left(x-y\right)\left(3a+b\right)\)
c) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)
\(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)
\(=\left(x-y\right)\left[x\left(x^2-2xy+y^2\right)-yx+y^2-y^2\right]\)
\(=\left(x-y\right)\left(x^3-2x^2y+xy^2-xy\right)\)
d) Ta có: \(2ax^3+6ax^2+6ax+18a\)
\(=2ax^2\left(x+3\right)+6a\left(x+3\right)\)
\(=\left(x+3\right)\left(2ax^3+6a\right)\)
\(=2a\left(x+3\right)\left(x^3+3\right)\)
e) Ta có: \(x^2y-xy^2-3x+3y\)
\(=xy\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-3\right)\)
d)
$x^4+2x^3+2x^2+2x+1$
$=(x^4+2x^3+x^2)+(x^2+2x+1)$
$=(x^2+x)^2+(x+1)^2=x^2(x+1)^2+(x+1)^2$
$=(x+1)^2(x^2+1)$
e)
$x^2y+xy^2+x^2z+y^2z+2xyz$
$=xy(x+y)+z(x^2+y^2)+2xyz$
$=xy(x+y)+z(x^2+y^2+2xy)$
$=xy(x+y)+z(x+y)^2=(x+y)(xy+zx+zy)$
f)
$x^5+x^4+x^3+x^2+x+1$
$=(x^5+x^4)+(x^3+x^2)+(x+1)=x^4(x+1)+x^2(x+1)+(x+1)$
$=(x+1)(x^4+x^2+1)$
$=(x+1)[(x^4+2x^2+1)-x^2]$
$=(x+1)[(x^2+1)^2-x^2]=(x+1)(x^2+1-x)(x^2+1+x)$
a)
$x^4-2x^3+2x-1=(x^4-2x^3+x^2)-(x^2-2x+1)$
$=(x^2-x)^2-(x-1)^2$
$=x^2(x-1)^2-(x-1)^2=(x-1)^2(x^2-1)=(x-1)^2(x-1)(x+1)$
$=(x-1)^3(x+1)$
b)
$a^6-a^4+2a^3+2a^2$
$=a^4(a^2-1)+2a^2(a+1)$
$=a^4(a-1)(a+1)+2a^2(a+1)$
$=(a+1)[a^4(a-1)+2a^2]$
$=a^2(a+1)[a^2(a-1)+2]$
$=a^2(a+1)(a^3-a^2+2)=a^2(a+1)[a^2(a+1)-2(a^2-1)]$
$=a^2(a+1)[a^2(a+1)-2(a-1)(a+1)]$
$=a^2(a+1)(a+1)(a^2-2a+2)=a^2(a+1)^2(a^2-2a+2)$
c)
$x^4+x^3+2x^2+x+1$
$=(x^4+2x^2+1)+(x^3+x)$
$=(x^2+1)^2+x(x^2+1)=(x^2+1)(x^2+1+x)$
Lời giải:
a)
$8^3:(-8)^{-5}=8^3.(-8)^5=8^3.(-8^5)=-8^3.8^5=-8^{3+5}=-8^{13}$
b)
$x^3y^4:(x^3y)=x^{3-3}.y^{4-1}=x^0.y^3=y^3$
c)
$5x^2y^4:(10x^2y)=(5:10).(x^2:x^2)(y^4:y)=\frac{1}{2}.1.y^3=\frac{1}{2}y^3$
d)
$\frac{3}{4}(xy)^3:(\frac{-1}{2}x^2y^2)$
$=(\frac{3}{4}: \frac{-1}{2})(x^3:x^2).(y^3:y^2)$
$=\frac{-3}{2}xy$