Rút gọn phân thức
A= \(\dfrac{\left(x^2-y\right)\left(y+1\right)+x^2y^2-1}{\left(x^2+y\right)\left(y+1\right)+x^2y^2+1}\)
Rút gọn phân thức:
\(a,\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(b,\dfrac{\left(x^2-y\right)\left(y+1\right)+x^2y^2-1}{\left(x^2+y\right)\left(y+1\right)+x^2y^2+1}\)
Rút gọn các biểu thức sau:
a/ \(\left(x-2y^{ }\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
b/ \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)
\(=2x^2-4xy+\dfrac{15}{4}y^2\)
b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(=2x^2+2x+13-2x^2+2\)
=2x+15
a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)
b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)
\(=2x+15\)
a; \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
= \(x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)
= \(2x^2-4xy+\dfrac{15}{4}y^2\)
b; \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
= \(x^2-4x+4+x^2+6x+9-2x^2+2\)
= \(2x+15\)
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:
\(a,\left(\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\right):\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(b,\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y-x}\right):\dfrac{2y}{x-y}\)
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
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
\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)
\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)
Rút gọn các phân thức sau:
a) \(\dfrac{6x^2y^2}{8xy^{ }5}\)
b) \(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
c) \(\dfrac{2x^2+2x
}{x+1}\)
d) \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
e) \(\dfrac{36\left(x-2\right)^3}{32-16x}\)
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
Tìm tập xác định, rồi rút gọn biểu thức:
B = \(\dfrac{y-x}{xy}\) : [\(\dfrac{y^2}{\left(x-y\right)^2\left(x+y\right)}\) - \(\dfrac{2x^2y}{x^4-2x^2y^2+y^4}\) + \(\dfrac{x^2}{\left(y^2-x^2\right)\left(x+y\right)}\)]
Tính giá trị của B với x = -\(\dfrac{1}{2}\), y = 2
Rút gọn phân thức:
\(a,\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(b,\dfrac{\left(x^2-y\right)\left(y+1\right)+x^2y^2-1}{\left(x^2+y\right)\left(y+1\right)+x^2y^2+1}\)
Câu a:
Xét tử số:
\(x^3-y^3+z^3+3xyz=(x-y)^3+3xy(x-y)+z^3+3xyz\)
\(=(x-y)^3+z^3+3xy(x-y+z)\)
\(=(x-y+z)[(x-y)^2-z(x-y)+z^2]+3xy(x-y+z)\)
\(=(x-y+z)(x^2+y^2+z^2-2xy-xz+yz)+3xy(x-y+z)\)
\(=(x-y+z)(x^2+y^2+z^2+xy+yz-xz)\)
Xét mẫu số:
\((x+y)^2+(y+z)^2+(z-x)^2\)
\(x^2+2xy+y^2+y^2+2yz+z^2+z^2-2zx+x^2\)
\(2(x^2+y^2+z^2+xy+yz-xz)\)
Do đó: \(\frac{x^3-y^3+z^3+3xyz}{(x+y)^2+(y+z)^2+(z-x)^2}=\frac{x-y+z}{2}\)
Câu b:
Xét tử số:
\((x^2-y)(y+1)+x^2y^2-1\)
\(=x^2y+x^2-y^2-y+x^2y^2-1\)
\(=(x^2y-y)+(x^2-1)+(x^2y^2-y^2)\)
\(=y(x^2-1)+(x^2-1)+y^2(x^2-1)=(x^2-1)(y^2+y+1)\)
Xét mẫu số:
\((x^2+y)(y+1)+x^2y^2+1\)
\(=x^2y+x^2+y^2+y+x^2y^2+1\)
\(=(x^2y+y)+(x^2+1)+(x^2y^2+y^2)\)
\(=y(x^2+1)+(x^2+1)+y^2(x^2+1)\)
\(=(x^2+1)(y+1+y^2)\)
Do đó:
\(\frac{(x^2-y)(y+1)+x^2y^2-1}{(x^2+y)(y+1)+x^2y^2+1}=\frac{(x^2-1)(y^2+y+1)}{(x^2+1)(y^2+y+1)}=\frac{x^2-1}{x^2+1}\)
phân tích đa thức \(\dfrac{1}{2}x^2-2y^2\) thành nhân tử
a. \(\dfrac{1}{2}x^2-2y^2=\dfrac{1}{2}\left(x^2-4y^2\right)=\dfrac{1}{2}\left(x-2y\right)\left(x+2y\right)\)
b. \(\dfrac{1}{2}x^2-2y^2=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\)
Cách phân tích nào đúng, a hay b ?