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Lời giải:
a.
\(-16a^4b^6-24a^5b^5-9a^6b^4=-[(4a^2b^3)^2+2.(4a^2b^3).(3a^3b^2)+(3a^3b^2)^2]\)
\(=-(4a^2b^3+3a^3b^2)^2=-[a^2b^2(4b+3a)]^2\)
\(=-a^4b^4(3a+4b)^2\)
b.
$x^3-6x^2y+12xy^2-8x^3$
$=x^3-3.x^2.2y+3.x(2y)^2-(2y)^3=(x-2y)^3$
c.
$x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}$
$=x^3+3.x^2.\frac{1}{2}+3.x.\frac{1}{2^2}+(\frac{1}{2})^3$
$=(x+\frac{1}{2})^3$
a) Ta có: \(-16a^4b^6-24a^5b^5-9a^6b^4\)
\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)
\(=-a^4b^4\cdot\left(4b+3a\right)^2\)
b) Ta có: \(x^3-6x^2y+12xy^2-8y^3\)
\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(x-2y\right)^3\)
c) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)
\(=x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)
\(=\left(x+\dfrac{1}{2}\right)^3\)
a/ -8x2 - 12xy3 - 4xy2 = -4x (2x + 3y3 + y2)
b/ (x - y)3 - x3 + y3 = x3 - 3x2y + 3xy2 - y3 - x3 + y3 = -3x2y + 3xy2 = -3xy (x - y)
c/ 3x(x - y) - 2y(x - y) = (x - y) (3x - 2y)
d/ a(x + 2) - 2b(x + 2) + (x + 2) = (x + 2) (a - 2b +1)
e/ 5x(x - y) - (y - x) = 5x(x - y) + (x - y) = (x - y) (5x + 1)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a: \(15x^2-5x^3=5x^2\left(3-x\right)\)
b: \(8x^3-y^3+4x^2y-2xy^2\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+2xy\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+4xy+y^2\right)\)
\(=\left(2x-y\right)\left(2x+y\right)^2\)
c: Ta có: \(x^8+64y^4\)
\(=x^8+16x^4y^2+64y^4-16x^4y^2\)
\(=\left(x^4+8y^2\right)^2-\left(4x^2y\right)^2\)
\(=\left(x^2-4x^2y+8y^2\right)\left(x^2+4x^2y+8y^2\right)\)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
a) Ta có: \(8x+4x^2-12xy\)
\(=4x\left(2+x-3y\right)\)
b) Ta có: \(5x^3-10x^2+5x\)
\(=5x\left(x^2-2x+1\right)\)
\(=5x\left(x-1\right)^2\)
c) Ta có: \(x^3+x^2y-xy^2-y^3\)
\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
d) Ta có: \(x^2-8x-9\)
\(=x^2-9x+x-9\)
\(=\left(x-9\right)\left(x+1\right)\)
a. `8x+4x^2-12xy=4x(2+x-3y)`
b) `5x^3-10x^2+5x=5x(x^2-2x+1)`
c) `x^3+x^2y-xy^2-y^3=x^2(x+y)-y^2(x+y)=(x+y)(x^2-y^2)=(x+y)^2 (x-y)`
d) `x^2-8x-9=(x^2-2.x.4+4^2)-25=(x-4)^2-5^2=(x+1)(x-9)`
a) x4 - x5 = x4( x - 1 )
b) -8x2y2 - 12xy3 - 4xy2
= -4xy( 2xy + 3y2 + y )
c) ( x - y )3 - x3 + y3
= x3 - 3x2y + 3xy2 - y3 - x3 + y3
= 3xy2 - 3x2y
= 3xy( y - x )