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a) 5x - 15y = 5(x - 3y)
b) \(\dfrac{3}{5}\)x2 + 5x4 - x2 - y
= \(\dfrac{3}{5}\)x2 + 5x2.x2 - x2 - y
= x2(\(\dfrac{3}{5}\) + 5x2 -1) - y
c) 14x2y2 - 21xy2 + 28x2y
= 7xy.xy - 7xy.3y + 7xy.4x
= 7xy(xy - 3y + 4x)
= 7xy[(xy - 3y) + 4x]
= 7xy[y(x - 3) +4x]
d) \(\dfrac{2}{7}x\)(3y - 1) - \(\dfrac{2}{7}y\)(3y - 1)
= (3y - 1).(\(\dfrac{2}{7}x\) - \(\dfrac{2}{7}y\) )
= (3y - 1).[\(\dfrac{2}{7}\)(x - y)]
e) x3 - 3x2 + 3x - 1
= x2.x - 3x.x + 3.x - 1
= x(x2-3x+3) - 1
g) 27x3 + \(\dfrac{1}{8}\)
= (3x)3 + \(\left(\dfrac{1}{2}\right)^3\)
= (3x + \(\dfrac{1}{2}\)).(9x2 - \(\dfrac{3}{2}\)x + \(\dfrac{1}{4}\))
h) (x+y)3 - (x-y)3
= 2(3x2y) + 2y3
f) (x+y)2 - 4x2
= -3x2 + y(2x + y)
\(a,\left(x+2y\right)^2=x^2+4xy+4y^2\)
\(b,\left(3x-2y\right)^2=9x^2-12xy+4y^2\)
\(c,\left(2x-\dfrac{1}{2}\right)^3=8x^3-3.4x^2.\dfrac{1}{2}+3.2x.\dfrac{1}{4}-\dfrac{1}{8}=8x^3-6x^2+\dfrac{3}{2}x-\dfrac{1}{8}\)
\(d,\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}+y\right)=\dfrac{x^2}{4}-y^2\)
\(2;a,x^4+4x^2+4\)
\(=\left(x^2+2\right)^2\)
\(b,4a^2b^2-c^2d^2\)
\(=\left(2ab\right)^2-\left(cd\right)^2\)
\(=\left(2ab-cd\right)\left(2ab+cd\right)\)
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)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
\(e,\)
\(\left(\dfrac{1}{3}a^3b+\dfrac{1}{3}a^2b^2-\dfrac{1}{4}ab^3\right):5ab\)
\(=\dfrac{1}{15}a^2+\dfrac{1}{15}ab-\dfrac{1}{20}b^2\)
\(f,\)
\(\left(-\dfrac{2}{3}x^5y^2+\dfrac{3}{4}x^4y^3-\dfrac{4}{5}x^3y^4\right):6x^2y^2\)
\(=-\dfrac{1}{9}x^3+\dfrac{1}{8}x^2y-\dfrac{2}{15}xy^2\)
\(g,\)
\(\left(\dfrac{3}{4}a^6b^3+\dfrac{6}{5}a^3b^4-\dfrac{5}{10}ab^5\right):\left(\dfrac{3}{5}ab^3\right)\)
\(=\dfrac{5}{4}a^5+2a^2b-\dfrac{5}{6}b^2\)
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$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!
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 ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
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 ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)