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a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
\(e,\)
\(\left(\dfrac{1}{3}a^3b+\dfrac{1}{3}a^2b^2-\dfrac{1}{4}ab^3\right):5ab\)
\(=\dfrac{1}{15}a^2+\dfrac{1}{15}ab-\dfrac{1}{20}b^2\)
\(f,\)
\(\left(-\dfrac{2}{3}x^5y^2+\dfrac{3}{4}x^4y^3-\dfrac{4}{5}x^3y^4\right):6x^2y^2\)
\(=-\dfrac{1}{9}x^3+\dfrac{1}{8}x^2y-\dfrac{2}{15}xy^2\)
\(g,\)
\(\left(\dfrac{3}{4}a^6b^3+\dfrac{6}{5}a^3b^4-\dfrac{5}{10}ab^5\right):\left(\dfrac{3}{5}ab^3\right)\)
\(=\dfrac{5}{4}a^5+2a^2b-\dfrac{5}{6}b^2\)
a) (5x-y)2 = (5x)2 - 2.5x.y + y2 = 25x2 - 10xy +y2
b) (2x + y2 )3 = (2x)3 - 3.(2x)2.y2 + 3.2x.(y2)2 - (y2)3 = 8x3 - 12x2y2 + 6xy4 - y6
c) (x + \(\dfrac{1}{4}\))2 = x2 + 2.x.\(\dfrac{1}{4}\) + (\(\dfrac{1}{4}\))2 = x2 + \(\dfrac{1}{2}\).x + \(\dfrac{1}{16}\)
d) (\(\dfrac{2}{3}\)x2 - \(\dfrac{1}{2}\)y)3 = (\(\dfrac{2}{3}\)x2)3 - 3.(\(\dfrac{2}{3}\)x2)2. \(\dfrac{1}{2}\)y + 3.\(\dfrac{2}{3}\)x2. ( \(\dfrac{1}{2}\)y)2 - (\(\dfrac{1}{2}\)y)3
= \(\dfrac{8}{27}\)x6 - \(\dfrac{2}{3}\)x4y + \(\dfrac{1}{2}\)x2y2 - \(\dfrac{1}{8}\)y3
câu b) của bạn làm sai rồi. hằng đẳng thức số 4 là : (A+B)3= A3+3A2B+3AB2+B3
sửa lại:
(2x+y2)3=2x3+3.(2x)2.y2+3.2x.(y2)2+(y2)3=8x3+12x2y2+6xy4+y6
a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
\(a\)) \(-2,5ab\left(-2a^2+3b^2\right)=5a^3b-7,5ab^3\)
b) \(-2x^3\left(3x+0,5x^2-7x^3-2\right)\)
\(=-6x^4-1x^5+14x^6+4x^3\)
c/ \(\left(x^3-2x^2+3x-5\right)\left(-xy\right)\)
\(=-x^4y+2x^3y-3x^2y+5xy\)
d/ \(\left(-\dfrac{1}{2}x^2y\right)\left(3x^3-\dfrac{2}{7}x^2-\dfrac{4}{5}x+8\right)\)
\(=-\dfrac{3}{2}x^5y+\dfrac{1}{7}x^4y+\dfrac{2}{5}x^3y-4x^2y\)
\(-2,5ab\left(-2a^2+3b^2\right)=5a^3b-7,5ab^3\)
\(-2x^3\left(3x+0,5x^2-7x^3-2\right)=-6x^4-x^5+14x^6+4x^3\)
\(\left(x^3-2x^2+3x-5\right)\left(-xy\right)=-x^4y+2x^3y-3x^2y+5xy\)
\(\left(\dfrac{1}{2}x^2y\right)\left(3x^3-\dfrac{2}{7}x^2-\dfrac{4}{5}x+8\right)\)
\(=\dfrac{-3}{2}x^5y+\dfrac{1}{7}x^4y+\dfrac{2}{5}x^3y-4x^2y\)
\(\left(x+\dfrac{1}{3}\right)^3=x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}\)
\(\left(2x+y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
\(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y\right)^3=\dfrac{1}{8}x^6+\dfrac{1}{4}x^4y+\dfrac{1}{6}x^2y^2+y^3\)
\(\left(x+\dfrac{1}{3}\right)^3=x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}\)
\(\left(2x+y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
\(\left(\dfrac{1}{2}x^2+\dfrac{1}{2}y\right)^3=\dfrac{1}{8}x^6+\dfrac{3}{8}x^4y+\dfrac{3}{8}x^2y^2+\dfrac{1}{8}y^3\)