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Để tính các biểu thức trên, ta sẽ áp dụng quy tắc nhân đa thức.
a) 2xy(3x+1) = 6x^2y + 2xy
b) -6x^2y(4x-5) = -24x^3y + 30x^2y
c) -3x^2(4x^2y-6xy) = -12x^4y + 18x^3y
d) 1/2xy^2(2x+3) = xy^2 + 3/2xy^2
e) 8x^2y^2(1/4xy-1/2x^2) = 2xy - 4x^2y^2
f) 5x(x^2+3x+1) = 5x^3 + 15x^2 + 5x
g) -1/2x^2y(2xy+6) = -x^3y - 3x^2y
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a) (2xy+5)(4x^2+5): = 2xy * 4x^2 + 2xy * 5 + 5 * 4x^2 + 5 * 5 = 8x^3y + 10xy + 20x^2 + 25 b) (6xy+4)(2x^2+1): = 6xy * 2x^2 + 6xy * 1 + 4 * 2x^2 + 4 * 1 = 12x^3y + 6xy + 8x^2 + 4 c) (9x^2+4)(3x+5): = 9x^2 * 3x + 9x^2 * 5 + 4 * 3x + 4 * 5 = 27x^3 + 45x^2 + 12x + 20 d) (-2xy+6)(1/2xy+7): = -2xy * 1/2xy + (-2xy) * 7 + 6 * 1/2xy + 6 * 7 = -xy + (-14xy) + 3 + 42 = -15xy + 45 e) (4x+1)(2x^2+5x+2): = 4x * 2x^2 + 4x * 5x + 4x * 2 + 1 * 2x^2 + 1 * 5x + 1 * 2 = 8x^3 + 20x^2 + 8x + 2x^2 + 5x + 2 = 8x^3 + 22x^2 + 13x + 2 f) (2x^2y+3x)(2x+1): = 2x^2y * 2x + 2x^2y * 1 + 3x * 2x + 3x * 1 = 4x^3y + 2x^2y + 6x^2 + 3x g) (4xy+5x^2y)(2xy+6): = 4xy * 2xy + 4xy * 6 + 5x^2y * 2xy + 5x^2y * 6 = 8x^2y^2 + 24xy + 10x^3y + 30x^2y = 8x^2y^2 + 30x^2y + 24xy h) (-1/2x^2+6)(4xy+5): = -1/2x^2 * 4xy + (-1/2x^2) * 5 + 6 * 4xy + 6 * 5 = -2xy + (-5/2x^2) + 24xy + 30 = 22xy + (-5/2x^2) + 30
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bai2 :cmr
a, a^3+b^3=(a+b)^3-3ab.(a+b)
VP= \(\left(a+b\right)^3-3ab\left(a+b\right)\)
=\(a^3+b^3+3a^2b+3ab^2-3a^2b-3ab^2=a^3+b^3\)
=VT
b.a^3-b^3=(a-b)^3+3ab,(a-b)
\(VP=\left(a-b\right)^3+3ab\left(a-b\right)\)
=\(a^3-3a^2b+ab^2.3-b^3+3a^2b-3ab^2=a^3-b^3\)
=VT
=> ĐPCM
bài 1.
a) = 8x^3+4x^2y+2xy^2-4x^2y-2xy^2-y^3-(8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3)
= 8x3+4x2y+2xy2-4x2y-2xy2-y3 - 8x3+4x2y-2xy2-4x2y+2xy2-y3
=-8x2y-6y3
b) = 27x3-18x2y+12xy2+18x2y-12xy2+8y3-27x3
=8y
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a)\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)
\(\Leftrightarrow\frac{\left(x+y\right)^2-1}{\left(x+1\right)^2-y^2}\)
\(\Leftrightarrow\frac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}\)
\(\Leftrightarrow\frac{x+y-1}{x-y+1}\)
b)\(\frac{3x^3-6x^2y+xy^2-2y^3}{9x^5-18x^4y-xy^4+2y^5}\)
\(\Leftrightarrow\frac{3x^2\left(x-2y\right)+y^2\left(x-2y\right)}{9x^4\left(x-2y\right)-y^4\left(x-2y\right)}\)
\(\Leftrightarrow\frac{\left(3x^2+y^2\right)\left(x-2y\right)}{\left(9x^4-y^4\right)\left(x-2y\right)}\)
\(\Leftrightarrow\frac{3x^2+y^2}{\left(3x^2-y^2\right)\left(3x^2+y^2\right)}\)
\(\Leftrightarrow\frac{1}{3x^2-y^2}\)
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sau bạn đăng tách ra cho mn cùng giúp nhé
a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)
b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)
c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)
d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)
e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)
f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)
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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)
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a,
$xy^2+x^2y+(-2xy^2)=xy^2-2xy^2+x^2y=-xy^2+x^2y$
b,
$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$
c,
$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$
d,
$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$
$=3x^2+x$
e,
$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$
f,
$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$
$=\frac{23}{4}xy^2$