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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)\(\dfrac{x^2}{x^2-y^2}\)-\(\dfrac{x-y}{x^2-y^2}\)
= \(\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\)= \(\dfrac{x}{x+y}\)
\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)
\(=2x^3+16x^2-5x\)
\(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)
\(=x\left(2x^2-1\right)+4x\left(4x-1\right)\left(ĐCCM\right)\)
a) Ta có: \(-3x^2\left(2x^2-\frac{1}{3}x+2\right)\)
\(=-6x^4+x^3-6x^2\)
b) Ta có: \(2xy^2\left(x-3y+xy\right)\)
\(=2x^2y^2-6xy^3+2x^2y^3\)
c) Ta có: \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-10x^2-4x^2+8x\)
\(=5x^3-14x^2+8x\)
d) Ta có: \(-\left(2-x\right)\left(2x+3\right)\)
\(=\left(x-2\right)\left(2x+3\right)\)
\(=2x^2+3x-4x-6\)
\(=2x^2-x-6\)
e) Ta có: \(\left(3x^3-2x^2+x\right):\left(-2x\right)\)
\(=\frac{-3}{2}x^2+x-\frac{1}{2}\)
f) Ta có: \(\left(15x^2y^2-21x^3y+2x^2y\right):\left(3x^2y\right)\)
\(=5y-7x+\frac{2}{3}\)
g) 
\(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)
\(=\left(2x-5\right).12\)
Những câu khác làm tương tự
a) x2 - 2x - 4y2 - 4y
= (x2 - 4y2) - (2x + 4y)
= (x + 2y)(x - 2y) - 2(x + 2y)
= (x + 2y)(x - 2y - 2)
= (x + 2y)[x - 2(y + 1)]
b) x4 + 2x3 - 4x - 4
= (x4 - 4) + ( 2x3 - 4x)
= (x2 - 2)(x2 + 2) + 2x(x2 - 2)
= (x2 - 2)(x2 + 2 + 2x)
c) x3 + 2x2y - x -2y
= (x3 - x) + (2x2y - 2y)
= x(x2 - 1) + 2y(x2 - 1)
= (x + 2y)(x2 - 1)
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+\(\dfrac{6}{5}\)
cảm ơn bạn nhé ^^