Question

[Trần Quang Minh]

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

Answer 1

[Trần Quang Minh]

Giúp mình nhé mọi người !

Answer 2

[Trần Nguyễn Bảo Quyên]

\(1.\)

\(a.\)

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

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

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

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

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

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

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

\(=x-1\)

\(b.\)

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

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

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

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

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

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

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

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

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