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27 tháng 11 2017

1) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\)

\(=\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{x-1}+1\)

\(=\dfrac{x^2}{x+1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x-1}+1\) MTC: \(\left(x-1\right)\left(x+1\right)\)

\(=\dfrac{x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^2\left(x-1\right)+2x+\left(x+1\right)+\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^3-x^2+2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\)

b) \(\dfrac{1}{x^3-x}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{x^2-1}\)

\(=\dfrac{1}{x\left(x^2-1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\) MTC: \(x\left(x-1\right)\left(x+1\right)\)

\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1-\left(x+1\right)+2x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1-x-1+2x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

11 tháng 12 2017

a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)

\(=\left(\dfrac{\left(2x+1\right)\left(2x+1\right)}{2x^2-1}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)

\(=\left(\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{2x^2-1}\right):\dfrac{4x}{10x-5}\)

\(=\left(\dfrac{\left(2x+1-2x-1\right)\left(2x+1+2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)

\(=\dfrac{4x}{2x^2-1}.\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{5}{2x+1}\)

b) \(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

\(=\left(\dfrac{1}{x^2+1}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+x-2\right)\)

\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)\)

\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{x^2-2x+1}{x}\right)\)

\(=\dfrac{\left(x-1\right)^2}{x^2+1}.\dfrac{x}{\left(x-1\right)^2}\)

\(=\dfrac{x}{x^2+1}\)

c) d) Tự làm đi mình làm biếng quass >.< ^^

24 tháng 6 2017

Phân thức đại số

Phân thức đại số

a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)

c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)

\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)

28 tháng 6 2017

Phép trừ các phân thức đại số

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,...
Đọc tiếp

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, \(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\)

2
8 tháng 12 2017

Giúp mình nhé mọi người ! leuleu

8 tháng 12 2017

\(1.\)

\(a.\)

\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=x-1\)

\(b.\)

\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)

\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)

\(=\dfrac{2y}{\left(x-y\right)}\)

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21 tháng 4 2017

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21 tháng 4 2017

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6 tháng 7 2018

ĐKXĐ: \(x\ne\pm y\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)}-\dfrac{2xy^2}{x^4-2x^2y^2+y^4}+\dfrac{7^2}{\left(x^2-y^2\right)\left(x+y\right)}\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)}-\dfrac{2xy^2}{\left(\left(x+y\right).\left(x-y\right)\right)^2}+\dfrac{49}{\left(x+y\right)^2.\left(x-y\right)}\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)^{ }}-\dfrac{2xy^2}{\left(x-y\right)^2.\left(x+y\right)^2}+\dfrac{49}{\left(x+y\right)^2.\left(x-y\right)}\)

\(A=\dfrac{x^2.\left(x+y\right)-2xy^2+49.\left(x-y\right)}{\left(x-y\right)^2.\left(x+y\right)^2}\)

\(A=\dfrac{x^3+x^2y-2xy^2+49x-49y}{\left(x-y\right)^2.\left(x+y\right)^2}\)

6 tháng 7 2018

ĐKXĐ: \(x\ne\pm1\)

\(B=\dfrac{x+3}{x+1}-\dfrac{2x-1}{x-1}-\dfrac{x-3}{x-1}\)

\(B=\dfrac{\left(x+3\right).\left(x-1\right)-\left(2x-1\right).\left(x+1\right)-\left(x-3\right)\left(x+1\right)}{\left(x+1\right).\left(x-1\right)}\)

\(B=\dfrac{x^2-x+3x-3-2x^2-2x+x+1-x^2-x+3x+3}{\left(x+1\right).\left(x-1\right)}\)

\(B=\dfrac{-4x^2+4x+1}{\left(x+1\right).\left(x-1\right)}=\dfrac{1+4x-4x^2}{\left(x+1\right).\left(x-1\right)}=\dfrac{\left(1-2x\right)^2}{\left(x+1\right).\left(x-1\right)}\)

29 tháng 6 2017

Biến đổi các biểu thức hữu tỉ. Giá trị của phân thức

Biến đổi các biểu thức hữu tỉ. Giá trị của phân thức