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a) x2(5x3 – x - \(\frac{1}{2}\)) = x2. 5x3 + x2 . (-x) + x2 . ( \(-\frac{1}{2}\) )
= 5x5 – x3 – \(\frac{1}{2}\)x2
b) (3xy – x2 + y) \(\frac{2}{3}\)x2y = \(\frac{2}{3}\)x2y . 3xy + \(\frac{2}{3}\)x2y . (- x2) + \(\frac{2}{3}\)x2y .
y = 2x3y2 – \(\frac{2}{3}\)x4y + \(\frac{2}{3}\)x2y2
c) (4x3– 5xy + 2x)( \(-\frac{1}{2}\)xy) = \(-\frac{1}{2}\)xy . 4x3 + ( \(-\frac{1}{2}\)xy) . (-5xy) + ( \(-\frac{1}{2}\)xy) . 2x
= -2x4y + \(\frac{5}{2}\)x2y2 – x2y.
a/ \(3x^2\left(4x^3-2x+\dfrac{1}{3}\right)=12x^5-6x^3+x^2\)
b/ \(\left(4x^2+8xy-3xy^2\right)\left(-\dfrac{3}{4}x^2y\right)\)
\(=-3x^4y-6x^3y^2+\dfrac{9}{4}x^3y^3\)
c/ \(4x^3\left(2x^2-x+5\right)5x=20x^4\left(2x^2-x+5\right)\)
\(=40x^6-20x^5+100x^4\)
a, \(3x^2\left(4x^3-2x+\dfrac{1}{3}\right)\)
\(=12x^5-6x^3+x^2\)
b, \(\left(4x^2+8xy-3xy^2\right).\left(\dfrac{-3}{4}x^2y\right)\)
\(=-3x^4y-6x^3y^2+\dfrac{9}{4}x^3y^3\)
c, \(4x^3\left(2x^2-x+5\right)5x\)
\(=\left(8x^5-4x^4+20x^3\right)5x\)
\(=40x^6-20x^5+100x^4=20x^4.\left(2x^2-x+5\right)\)
Chúc bạn học tốt!!! Mình không chắc đâu !
\(a,\left(4x^3-3xy^2+2xy\right).\left(-\dfrac{1}{3}x^2y\right)\)
\(=\dfrac{-x}{3y}+\dfrac{y}{x}-\dfrac{2}{3x}\)
\(b,\left(5xy-x^2+y\right)\left(\dfrac{2}{5}xy^2\right)\)
\(=\dfrac{2}{y}-\dfrac{2x}{5y^2}+\dfrac{2}{5xy}\)
c,=\(\dfrac{-4}{3}x^5y+x^3y^3-\dfrac{2}{3}x^3y^2\) b,=\(2x^2y^3-\dfrac{2}{5}x^3y^2+\dfrac{2}{5}xy^3\)
a) \(x^2+4y^2+4xy\)
\(=x^2+4xy+4y^2\)
\(=\left(x+2y\right)^2\)
\(a,x^2+4xy+4y^2=x^2+2.x.2y+\left(2y\right)^2.\)
\(=\left(x+2y\right)^2\)
\(b,\left(\frac{3}{2}x\right)^2-3xy+y^2\)
\(=\left(\frac{3}{2}x\right)^2-2.\frac{3}{2}x.y+y^2\)
\(=\left(\frac{3}{2}x-y\right)^2\)
\(c,\frac{x^2}{9}+\frac{x}{3}+\frac{1}{4}\)
\(=\left(\frac{x}{3}\right)^2+2.\frac{x}{3}.\frac{1}{2}+\left(\frac{1}{2}\right)^2\)
\(=\left(\frac{x}{3}+\frac{1}{2}\right)^2\)
hẳng đẳng thức tề
(a+b)^2= a^2+2ab+b^2
(a+b)^3= a^3+3a^2b+3ab^2+b^3
a^2-b^2= (a+b)(a-b)
a,\(\left(-\frac{1}{2}x+\frac{1}{4}y^2\right)^2=\left(-\frac{1}{2}x\right)^2+2\left(-\frac{1}{2}x\right).\left(\frac{1}{4}y^2\right)+\left(\frac{1}{4}y^2\right)^2\)
\(=\frac{1}{4}x^2-\frac{1}{4}xy^2+\frac{1}{16}y^4\)
b,\(\left(x+3xy\right)^3=x^3+3.x^2.3xy+3.x.\left(3xy\right)^2+\left(3xy\right)^3\)
\(=x^3+9x^3y+27x^3y^2+27x^3y^3\)
c, \(\left(-2\sqrt{2}+\sqrt{3}\right)^2-\left(\sqrt{3}+3\sqrt{2}\right)^2\)
\(=\left(-2\sqrt{2}\right)^2+2.\left(-2\sqrt{2}\right).\sqrt{3}+\sqrt{3}^2-\left[\sqrt{3}^2+2.3\sqrt{2}.\sqrt{3}+\left(3\sqrt{2}\right)^2\right]\)
\(=4.2-4.\sqrt{6}+3-3-6\sqrt{6}-9.2\)
\(=-10-10\sqrt{6}\)
) \(\dfrac{x^3+8y^3}{2y+x}\)
\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)
\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)
\(=x^2+2xy+4y^2\)
b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)
\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)
\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)
\(=\dfrac{3a-1}{2\left(a-4\right)}\)
c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)
\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2}\)
d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)
\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)
\(=x^2-10x+25+7x+14-x^2-2x\)
\(=39-5x\)
e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)
\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)
\(=\dfrac{3x+2x+1}{x-2}\)
\(=\dfrac{5x+1}{x-2}\)
h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$b) $ \(\left(-x\right)^7:\left(-x\right)^5\)
\(=\left(-x\right)^{7-5}\)
\(=\left(-x\right)^2\)
\(=x^2\)
$c)$ \(5x^2y^4:10x^2y\)
\(=\dfrac{1}{2}y^3\)
$e)$ \(\left(-xy\right)^{14}:\left(-xy\right)^7\)
\(=\left(-xy\right)^{14-7}\)
\(=\left(-xy\right)^7\)
Các câu còn lại tương tự nha bạn!
Lời giải:
\((4x^3-3xy^2+3xy)(-\frac{1}{3}x^2y)=4x^3.\frac{-1}{3}x^2y-3xy^2.-\frac{1}{3}x^2y+3xy.\frac{-1}{3}x^2y\)
\(=-\frac{4}{3}x^5y+x^3y^3-x^3y^2\)