Ta có: \(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\)
\(=\left(x^4-y^4\right)\left(x^4+y^4\right)\)
\(=x^8-y^8\)
Ta có: \(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\)
\(=\left(x^4-y^4\right)\left(x^4+y^4\right)\)
\(=x^8-y^8\)
Bài 4: Chứng minh
\(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
a) \(\dfrac{1}{\left(x-y\right)\left(y-z\right)}+\dfrac{1}{\left(y-z\right)\left(z-x\right)}+\dfrac{1}{\left(z-x\right)\left(x-y\right)}\)
b) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-z\right)\left(y-x\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}\)
c) \(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-x\right)\left(y-z\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
cho x + y + z = 0 và x, y , z khác 0 hãy rút gọn
a) P = \(\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b) Q = \(\dfrac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
\(\dfrac{x}{\left(x+y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
Chứng minh:
\(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
Rút gọn phân thức:\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
Cho biểu thức \(A=\dfrac{\left(x^2+y\right)\left(y+\dfrac{1}{4}\right)+x^2y^2+\dfrac{3}{4}\left(y+\dfrac{1}{3}\right)}{x^2y^2+1+\left(x^2-y\right)\left(1-y\right)}\)
a) CMR: Biểu thức A không phụ thuộc vào biến \(x\) ?
b) Tìm Min A ?
Bài 1: Thực hiện phép tính
a, \(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}\)+\(\dfrac{2}{x^2+3}\)+\(\dfrac{1}{x+1}\)
b, \(\dfrac{x+y}{2\left(x-y\right)}\)-\(\dfrac{x-y}{2\left(x+y\right)}\)+\(\dfrac{2y^2}{x^2-y^2}\)
c, \(\dfrac{x-1}{x^3}\)-\(\dfrac{x+1}{x^3-x^2}\)+\(\dfrac{3}{x^3-2x^2+x}\)
d, \(\dfrac{xy}{ab}\)+\(\dfrac{\left(x-a\right)\left(y-a\right)}{a\left(a-b\right)}\)-\(\dfrac{\left(x-b\right)\left(y-b\right)}{b\left(a-b\right)}\)
e, \(\dfrac{x^3}{x-1}\)-\(\dfrac{x^2}{x+1}\)-\(\dfrac{1}{x-1}\)+\(\dfrac{1}{x+1}\)
f, \(\dfrac{x^3+x^2-2x-20}{x^2-4}\)-\(\dfrac{5}{x+2}\)+\(\dfrac{3}{x-2}\)
g, \(\left\{\dfrac{x-y}{x+y}+\dfrac{x+y}{x-y}\right\}\).\(\left\{\dfrac{x^2+y^2}{2xy}\right\}\).\(\dfrac{xy}{x^2+y^2}\)
h, \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}\)+\(\dfrac{1}{\left(b-c\right)\left(c-a\right)}\)+\(\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
i, \(\dfrac{\left[a^2-\left(b+c\right)^2\right]\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
k, \(\left[\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left\{\dfrac{x^2}{y}-\dfrac{y^2}{x}\right\}\right]\):\(\dfrac{x-y}{x}\)
Bài 2: Rút gọn các phân thức:
a, \(\dfrac{25x^2-20x+4}{25x^2-4}\)
b, \(\dfrac{5x^2+10xy+5y^2}{3x^3+3y^3}\)
c, \(\dfrac{x^2-1}{x^3-x^2-x+1}\)
d, \(\dfrac{x^3+x^2-4x-4}{x^4-16}\)
e, \(\dfrac{4x^4-20x^3+13x^2+30x+9}{\left(4x^2-1\right)^2}\)
Bài 3: Rút gọn rồi tính giá trị các biểu thức:
a, \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\) với a = 4, b = -5, c = 6
b, \(\dfrac{16x^2-40xy}{8x^2-24xy}\) với \(\dfrac{x}{y}\) = \(\dfrac{10}{3}\)
c, \(\dfrac{\dfrac{x^2+xy+y^2}{x+y}-\dfrac{x^2-xy+y^2}{x-y}}{x-y-\dfrac{x^2}{x+y}}\) với x = 9, y = 10
Bài 4: Tìm các giá trị nguyên của biến số x để biểu thức đã cho cũng có giá trị nguyên:
a, \(\dfrac{x^3-x^2+2}{x-1}\)
b, \(\dfrac{x^3-2x^2+4}{x-2}\)
c, \(\dfrac{2x^3+x^2+2x+2}{2x+1}\)
d, \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\)
e, \(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\)
Chứng minh các đẳng thức sau:
a, \(\frac{3x}{x+y}=\frac{-3x\left(x-y\right)}{y^2-x^2}\left(x\ne-y,x\ne y\right)\)
b, \(\frac{x-2}{-x}=\frac{8xy^2}{12ay}\left(a\ne0,y\ne0\right)\)
c, \(\frac{x+y}{3a}=\frac{3a\left(x+y\right)^2}{9a^2\left(x+y\right)}\left(a\ne0,x\ne-y\right)\)