So sánh 2 biểu thức sau A=\(\dfrac{y^2\left(x+1\right)+\left(x+1\right)}{y^2+1}\) và B=\(\dfrac{y^2\left(x-1\right)+2x-x}{y^2+2}\)
Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
Rút gọn các biểu thức sau :
a)\(\dfrac{25xy^3\left(2x-y\right)^2}{75xy^2\left(y-2x\right)}\)
b)\(\dfrac{x^2-y^2}{x^2-y^2+xz-yz}\)
c)\(\dfrac{\left(2x+3\right)-x^2}{x^2-1}\)
d)\(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}\)
`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`
`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`
`=(y(y-2x))/3`
`b,(x^2-y^2)/(x^2-y^2+xz-yz)`
`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`
`=(x+y)/(x+y+z)`
`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`
`=(-(x^2-3x+x-3))/((x-1)(x+1))`
`=(-x(x-3)+x-3)/((x-1)(x+1))`
`=((x-3)(1-x))/((x-1)(x+1))`
`=(3-x)/(1+x)`
`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`
`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`
`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`
`=(3x^2-4x+1)/(2x^2+x-3)`
`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`
`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`
`=(3x-1)/(2x+3)`
a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)
\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)
\(=\dfrac{y\left(y-2x\right)}{3}\)
A= \(\left(\dfrac{x+y}{y}-\dfrac{2y}{y-x}\right):\left(\dfrac{x^2+y^2}{y-x}\right)+\left(\dfrac{x^2+1}{2x-1}-\dfrac{x}{2}\right).\dfrac{1-2x}{x+2}\)
Với điều kiện của x, y để A có nghĩa, hãy rút gọn biểu thức trên
\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)
\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)
khai triển các biểu thức sau:
\(a.\left(2x+3y\right)^2\)
\(b.2\left(\dfrac{1}{2}x^2+y\right)\left(x^2-2y\right)\)
\(c.\left(x+y+z\right)^2\)
a. (2x+3y)2= (2x)2+2.2x.3y+(3y)2
=4x2+12xy+9y2
b. 2(\(\dfrac{1}{2}\)x2+y)(x2-2y)
=(x2+2y)(x2-2y)
=x4-4y2
c, (x+y+z)2= [(x+y)+z]2
=(x+y)2+2(x+y)z+z2
=x2+2xy+y2+2xz+2yz+z2
=x2+y2+z2+2xy+2yz+2xz
Rút gọn biểu thức:
\(a,\left(\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\right):\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(b,\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y-x}\right):\dfrac{2y}{x-y}\)
a: \(=\left(\dfrac{x}{y\left(x-y\right)}-\dfrac{2x-y}{x\left(x-y\right)}\right):\dfrac{x+y}{xy}\)
\(=\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}\cdot\dfrac{xy}{x+y}\)
\(=\dfrac{\left(x-y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x-y}{x+y}\)
b: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\)
\(=\dfrac{4xy+4y^2}{2\left(x+y\right)}\cdot\dfrac{1}{2y}=\dfrac{4y\left(x+y\right)}{4y\left(x+y\right)}=1\)
Rút gọn biểu thức:
\(a,\left(\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\right):\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(b,\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y-x}\right):\dfrac{2y}{x-y}\)
Rút gọn biểu thức:
\(a,\left(\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\right):\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(b,\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y-x}\right):\dfrac{2y}{x-y}\)
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
b2: so sánh hai biểu thức
\(A=\dfrac{y^2\left(x+1\right)+x+1}{y^2+1}\)
\(B=\dfrac{y^2\left(x-1\right)+2x-2}{x+y}\)
\(A=\dfrac{y^2\left(x+1\right)+\left(x+1\right)}{y^2+1}=\dfrac{\left(x+1\right)\left(y^2+1\right)}{y^2+1}=x+1\)
\(B=\dfrac{y^2\left(x-1\right)+2\left(x-1\right)}{x+y}=\dfrac{\left(x-1\right)\left(y^2+2\right)}{x+y}\)