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b: 6x^2-24y^2
=6(x^2-4y^2)
=6(x-2y)(x+2y)
c: =(4x)^3-(3y)^3
=(4x-3y)(16x^2+12xy+9y^2)
d: x^4-2x^3-x^2
=x^2(x^2-2x-1)
a: =64x^4+16x^2y^2+y^4-16x^2y^2
=(8x^2+y^2)^2-(4xy)^2
=(8x^2+y^2-4xy)(8x^2+y^2+4xy)
b: =x^8+2x^4+1-x^4
=(x^4+1)^2-x^4
=(x^4-x^2+1)(x^4+x^2+1)
=(x^4-x^2+1)(x^4+2x^2+1-x^2)
=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)
c: =(x+1)(x^2-x+1)+2x(x+1)
=(x+1)(x^2-x+1+2x)
=(x+1)(x^2+x+1)
d: =(x^2-1)(x^2+1)-2x(x^2-1)
=(x^2-1)(x^2-2x+1)
=(x-1)^2*(x-1)(x+1)
=(x+1)(x-1)^3
Lời giải:
a.
$64x^2-24y^2=8(8x^2-3y^2)=8(\sqrt{8}x-\sqrt{3}y)(\sqrt{8}x+\sqrt{3}y)$
b.
$64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)$
c.
$x^4-2x^3+x^2=(x^2-x)^2=[x(x-1)]^2=x^2(x-1)^2$
d.
$(x-y)^3+8y^3=(x-y)^3+(2y)^3=(x-y+2y)[(x-y)^2-2y(x-y)+(2y)^2]$
$=(x+y)(x^2-4xy+7y^2)$
a) \(64x^2-24y^2\)
\(=8\left(8x^2-3y^2\right)\)
b) \(64x^3-27y^3\)
\(=\left(4x\right)^3-\left(3y\right)^3\)
\(=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
c) \(x^4-2x^3+x^2\)
\(=x^2\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)^2\)
d) \(\left(x-y\right)^3+8y^3\)
\(=\left(x-y+2y\right)\left(x^2-2xy+y^2-2xy+2y^2+4y^2\right)\)
\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)
x2 + 1 - y2 - 2x
= x2 - 2x + 1 - y2
=[x2 - 2x + 1] - y2
=[x-1]2 - y2
=[x-1-y][x-1+y]
a) \(x^2+1-y^2-2x=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
b) \(64x^4+y^4=\left(8x^2\right)^2+\left(y^2\right)^2=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
Lời giải:
a. Không phân tích được nữa
b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$
$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$
64x4 + y4 = (8x2)2 +16x2y2+ (y2) - 16x2y2 = (8x2+y2)2 - (4xy)2 = (8x2+y2- 4xy) (8x2+y2 + 4xy)
mk chỉ hơi chửi tục tí thôi nhưng địt con mẹ mình hiền lắm
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(64x^4+y^4=64x^4+16x^2y^2-16x^2y^2+y^4\)
\(=\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
\(3,x\left(x-1\right)-y\left(1-x\right)=\left(x+y\right)\left(x-1\right)\\ 4,x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\\ 5,x^2-2xy+y^2-xz+yz=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y-z\right)\left(x-y\right)\\ 6,x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\\ 9,x^3+x^2-xy+xy+y^2+y^3\\ =x^2\left(x+1\right)+y^2\left(x+1\right)=\left(x^2+y^2\right)\left(x+1\right)\\ 10,x^2-6\left(x+3\right)-9\\ =x^2-6x-18-9\\ =x^2-6x-27=\left(x-9\right)\left(x+3\right)\)
10: \(x^2-6\left(x+3\right)-9\)
\(=x^2-6x-18-9\)
\(=x^2-6x-27\)
\(=\left(x-9\right)\left(x+3\right)\)