Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
a/ (-5x3)(2x2+3x-5)
=-10x5-15x4+25x3
b/(2x-1)x
=2x2-x
c/(x-y)(3x2+4xy)
=3x3+4x2y-3x2y-4xy2
=3x3 +x2y-4xy2
Câu 2:
a/ x3-2x2+x
=x(x2-2x+1)
=x(x-1)2
b/x2-x-12
=x2 +3x-4x-12
=(x2 +3x)+(-4x-12)
=x(x+3)-4(x+3)
=(x+3)(x-4)
c/ 2x-6
=2(x-3)
e/ x2+4x+4-y2
=(x2+4x+4)-y2
=(x+2)2-y2
=(x+2-y)(x+2+y)
d/ x2-2xy+y2-16
=(x2-2xy+y2)-16
=(x-y)2-16
=(x-y-4)(x-y+4)
Câu 3:
a: \(=\dfrac{5xy-4+3xy+4}{2x^2y^3}=\dfrac{8xy}{2x^2y^3}=\dfrac{4}{xy^2}\)
b: \(=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)
\(=\dfrac{y^2-12y+36}{6y\left(y-6\right)}=\dfrac{y-6}{6y}\)
c: \(=\dfrac{3x+1-2x+3}{x+y}=\dfrac{x+4}{x+y}\)
d: \(=\dfrac{4x+7+5x+7}{9}=\dfrac{9x+14}{9}\)
e: \(=\dfrac{5\left(x+2\right)}{2\left(2x-1\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-5\left(x-2\right)}{2x-1}\)
1) 4x\(^2\).(5x3+2x-1)
= 20x\(^5\)+8x\(^3\)-4x\(^2\).
2) 4x\(^3\): x2
= 4x
3) ( 15x2y3-10x3y3+6xy): 5xy
= 3xy2-2x2y2+\(\dfrac{6}{5}\)
4) (5x3+14x2+12x+8 ): (x+2)
= 5x2+4x+4
5)\(\dfrac{7}{2x}\)+\(\dfrac{11}{3y^2}\)
=\(\dfrac{7.3y^2+11.2x}{6xy^2}\) =\(\dfrac{21y^2+22x}{6xy^2}\) = \(\dfrac{21+22}{6}\) =\(\dfrac{43}{6}\)
6) \(\dfrac{x}{x+2}\) +\(\dfrac{3}{\left(x+2\right)\left(4x-7\right)}\)
7)\(\dfrac{3}{x-y}\)-\(\dfrac{2x^2}{x+y}\)
= \(\dfrac{3\left(x+y\right)-2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{3x+3y-2x-2y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{1}{x-y}\).
8)\(\dfrac{1}{2}\)x2y2.(2x+y)(2x-y)
= \(\dfrac{1}{2}\)x2y2.(4x2-2xy+2xy-y2)
= \(\dfrac{1}{2}\)x2y2.(4x2-y2)
= 2x4y2-\(\dfrac{1}{2}\)x2y4
9) (x-\(\dfrac{1}{2}\)).(x+\(\dfrac{1}{2}\)).(4x-1)
= x2.(4x-1)
= 4x3-x2
10)\(\dfrac{3x}{2x+6}\)+\(\dfrac{6-x}{2x^2+6x}\)
= \(\dfrac{3x}{2\left(x+3\right)}\)+\(\dfrac{6-x}{2x\left(x+3\right)}\)= \(\dfrac{3x^2+6-x}{2x\left(x+3\right)}\)=\(\dfrac{3-x}{3}\)= -x
11) x2-\(\dfrac{1}{2x-2}\)+3x+\(\dfrac{3}{1-x^2}\)
12)\(\dfrac{x^2}{x^2-y^2}\)-\(\dfrac{x-y}{x^2-y^2}\)
= \(\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\)= \(\dfrac{x}{x+y}\)
Bài 1:
a) \(x^2-y^2+10x+25\)
\(=\left(x^2+10x+25\right)-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+y+5\right)\left(x-y+5\right)\)
b) \(x^3-x^2-5x+125\)
\(=x^3+5x^2-6x^2-30x+25x+125\)
\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
c) \(x^4+4y^4\)
\(=\left(x^2\right)^2+2x^22y^2+\left(2y^2\right)^2-2x^22y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)
d)Sửa đề \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left[\left(b^2-c^2\right)+\left(a^2-b^2\right)\right]+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)-b\left(a^2-b^2\right)+c\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b^2-c^2\right)-\left(b-c\right)\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c\right)-\left(b-c\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c-a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
e) \(7x^2-10xy+3y^2\)
\(=\left(\sqrt{7}x\right)^2-2.\sqrt{7}x.\sqrt{3}y+\left(\sqrt{3}y\right)^2\)
\(=\left(\sqrt{7}x-\sqrt{3}y\right)^2\)
f) Sửa đề \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
h) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=x^2y+xy^2-y^2z-yz^2+x^2z-xz^2\)
\(=\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y^2-z^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y+z\right)\left(y-z\right)-yz\left(y+z\right)\)
\(=\left(y+z\right)\left[x^2+x\left(y-z\right)-yz\right]\)
\(=\left(y+z\right)\left(x^2+xy-xz-yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)-z\left(x+y\right)\right]\)
\(=\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
\(=x^3+x^2-\left(4x+4\right)=x^2\left(x+1\right)-4\left(x+1\right)=\left(x^2-4\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
\(x^4+x^3+x^2-1=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)=\left(x+1\right)\left(x^3+x-1\right)\)
\(c,=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
\(d,=x^2y^2-y^2-x^2+1=\left(x^2-1\right)\left(y^2-1\right)=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
\(e,4x^2+4x-15=\left(4x^2+4x+1\right)-16=\left(2x+1\right)^2-4^2=\left(2x+5\right)\left(2x-3\right)\)
\(3x^2-7x+2=\left(3x^2-6x\right)-\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)=\left(3x-1\right)\left(x-2\right)\)
\(4x^2-5x+1=\left(4x^2-4x\right)-\left(x-1\right)=4x\left(x-1\right)-\left(x-1\right)=\left(4x-1\right)\left(x-1\right)\)
Phân tích à :v
a) x3 + x2 - 4x - 4 = x2( x + 1 ) - 4( x + 1 ) = ( x + 1 )( x2 - 4 ) = ( x + 1 )( x - 2 )( x + 2 )
b) x4 + x3 + x2 - 1 = x3( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 )
c) x2 + 2xy + y2 - 2x - 2y + 1 = ( x2 + 2xy + y2 ) - ( 2x + 2y ) + 1 = ( x + y )2 - 2( x + y ) + 12 = ( x + y - 1 )2
d) x2y2 + 1 - x2 - y2 = ( x2y2 - x2 ) - ( y2 - 1 ) = x2( y2 - 1 ) - ( y2 - 1 ) = ( y2 - 1 )( x2 - 1 ) = ( y - 1 )( y + 1 )( x - 1 )( x + 1 )
e) 4x2 + 4x - 15 = ( 4x2 + 4x + 1 ) - 16 = ( 2x + 1 )2 - 42 = ( 2x + 1 - 4 )( 2x + 1 + 4 ) = ( 2x - 3 )( 2x + 5 )
g) 3x2 - 7x + 2 = 3x2 - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )
h) 4x2 - 5x + 1 = 4x2 - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )
a, \(\frac{3x^3y^4}{-5xy^2}=-\frac{3}{5}x^2y^2\)
b, \(\frac{-4x^3y^5}{-\frac{1}{2}x^3y}=8y^4\)
c, \(\frac{5x^2y^2+10x^2y^2}{-5x^2y^3}=\frac{15x^2y^2}{-5x^2y^3}=\varnothing\)( Vì y^2 \(⋮̸̸\)y^3 )