Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải
1) 3xy2 : 5x = \(\frac{3}{5}\)y2
2) 15x4yz3 : 4xyz = \(\frac{15}{4}\)x3z2
3) (4x2y2 - 12xy3 - 7x) : 3x = \(\frac{4}{3}\)xy2 - 4y3 - \(\frac{7}{3}\)
4) (14x4y2 - 12xy3 - x) : 4x = \(\frac{7}{2}\)x3y2 - 3y3 - \(\frac{1}{4}\)
5) (6x2 + 13x - 5) : (2x + 5) = (3x - 1)(2x + 5) : (2x + 5) = 3x - 1
6) (2x4 + x3 - 5x2 - 3x - 3) : (x2 - 3)
= 2x4 + x2 - 6x2 + x3 - 3 - 3x : x2 - 3
= x2(2x2 + x + 1) - 3(2x2 + x + 1) : x2 - 3
= (2x2 + x + 1)(x2 - 3) : x2 - 3
= 2x2 + x + 1
Bài 2:
a: \(A=\left(2x-y\right)^2=\left(12-2\right)^2=100\)
b: \(=\left(x-3\right)^3=100^3=1000000\)
c: \(=\left(x-y\right)^2-9z^2\)
\(=\left(x-y-3z\right)\left(x-y+3z\right)\)
\(=\left(6+4-90\right)\left(6+4+90\right)=-80\cdot100=-8000\)
Bài 1:
a)\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
b)\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
c)Đề sai hoàn toàn
d) \(2x^2+4xy+2y^2-8z^2=2\left(x^2+2xy+y^2-4z^2\right)=2\left[\left(x+y\right)^2-\left(2z\right)^2\right]=2\left(x+y-2z\right)\left(x+y+2z\right)\)e) \(3x-3a+yx-ya=3\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(3+y\right)\)
f)\(\left(x^2+y^2\right)^2-4x^2y^2=\left(x-y\right)^2\left(x+y\right)^2\)
g)\(2x^2-5x+2=2x^2-x-4x+2=x\left(2x-1\right)-2\left(2x-1\right)=\left(2x-1\right)\left(x-2\right)\)
i)\(14x\left(x-y\right)-21y\left(y-x\right)+28z\left(x-y\right)=14x\left(x-y\right)+21y\left(x-y\right)+28z\left(x-y\right)=7\left(x-y\right)\left(2x+3y+4z\right)\)
1,4x2.(5x3+2x-1)
=4x2.5x3+4x2.2x-4x2.1
20x5+8x3-4x2
2,4x3y2:x2
=4xy2
3,(15x2y3-10x3y3+6xy):5xy
15x2y3:5xy-10x3y3:5xy+6xy:5xy
3xy2-2x2y2+\(\dfrac{6}{5}\)
1: \(=20x^5+8x^3-4x^2\)
2: \(=4xy^2\)
3: \(=3xy^2-2x^2y^2+\dfrac{6}{5}\)
4: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
5: \(=\dfrac{7}{2x}+\dfrac{11}{3y^2}=\dfrac{21y^2+22x}{6xy^2}\)
6: \(=\dfrac{4x^2-7x+3}{\left(4x-7\right)\left(x+2\right)}\)
7: \(=\dfrac{3x+3y-2x^3+2x^2y}{\left(x-y\right)\left(x+y\right)}\)
8: \(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\dfrac{1}{2}x^2y^4\)
9: \(=\left(x-\dfrac{1}{4}\right)\left(4x-1\right)=4\left(x-\dfrac{1}{4}\right)^2=4\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)
\(=4x^2-2x+\dfrac{1}{4}\)
10: \(=\dfrac{3x^2+6-x}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
11: \(=\dfrac{x+1}{2}-\dfrac{3}{x-1}\)
\(=\dfrac{x^2-7}{2\left(x-1\right)}\)
12: \(=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)
15:=x^3-y^3+2
1: \(=20x^5+8x^3-4x^2\)
2: \(=4xy^2\)
3: \(=3xy^2-2x^2y^2+\dfrac{6}{5}\)
4: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
5: \(=\dfrac{7}{2x}+\dfrac{11}{3y^2}=\dfrac{21y^2+22x}{6xy^2}\)
6: \(=\dfrac{4x^2-7x+3}{\left(4x-7\right)\left(x+2\right)}\)
7: \(=\dfrac{3x+3y-2x^3+2x^2y}{\left(x-y\right)\left(x+y\right)}\)
8: \(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\dfrac{1}{2}x^2y^4\)
9: \(=\left(x-\dfrac{1}{4}\right)\left(4x-1\right)=4\left(x-\dfrac{1}{4}\right)^2=4\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)
\(=4x^2-2x+\dfrac{1}{4}\)
10: \(=\dfrac{3x^2+6-x}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
11: \(=\dfrac{x+1}{2}-\dfrac{3}{x-1}\)
\(=\dfrac{x^2-7}{2\left(x-1\right)}\)
12: \(=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)
15:=x^3-y^3+2
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 3 : Ta có : \(A=\frac{2}{5}xy\left(x^2y-5x+10y\right)\)
\(A=\frac{2}{5}xy\cdot x^2y+\frac{2}{5}xy\left(-5x\right)+\frac{2}{5}xy\cdot10y\)
\(A=\frac{2}{5}x^3y^2-2x^2y+4xy^2\)
Chọn C
Bài 4 : \(\left(x-2\right)\left(x+5\right)=x\left(x+5\right)-2\left(x+5\right)\)
\(=x^2+5x-2x-10\)
\(=x^2+3x-10\)
Chọn B
Bài 3 :
Ta có: A = 2/5xy( x2y -5x + 10y )
= 2/5xy.x2y - 2/5xy.5x + 2/5xy.10y
= 2/5x3y2 - 2x2y + 4xy2.
Chọn đáp án C
Bài 4 :
Ta có ( x - 2 )( x + 5 )
= x( x + 5 ) - 2( x + 5 )
= x2 + 5x - 2x - 10 = x2 + 3x - 10.
Chọn đáp án B.
Hok tốt
a: \(\dfrac{10x^3y-5x^2y^2-25x^4y^3}{-5xy}=-2x^2+xy+5x^3y^2\)
c: \(\dfrac{27x^3-y^3}{3x-y}=9x^2+3xy+y^2\)