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24 tháng 6 2017

Phân thức đại số

Phân thức đại số

21 tháng 4 2017

a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x

=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x

=8x.5(2x+1)(2x−1)(2

29 tháng 11 2018

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

4 tháng 12 2018

a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)

( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)

(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)

1/x +1/x+4

2x+4/x(x+4)

4 tháng 12 2018

Câu b bạn tách các mẫu thành nhân tử rồi làm như câu a nhé

13 tháng 12 2018

\(\frac{x^2+3x+9}{2x+10}.\frac{x+5}{x^3-27}\)

\(=\frac{x^2+3x+9}{2\left(x+5\right)}.\frac{x+5}{\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\frac{\left(x+5\right)\left(x^2+3x+9\right)}{2\left(x+5\right)\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\frac{1}{2\left(x-3\right)}\)

\(\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\left(\frac{x^2-36}{x^2+1}\right)\)

\(=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\left[\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\right]\)

\(=\frac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12x^2+12}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12\left(x^2+1\right).\left(x-6\right)\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(x^2+1\right)}\)

\(=\frac{12}{x}\)

a: \(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)

\(=\dfrac{3}{x-3}\cdot\dfrac{-\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{-3}{x-3}\)

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

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

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

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

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

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

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

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

\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)

\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\dfrac{-\left(x^2-x-6\right)+x^2-1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}\)

\(=\dfrac{-x^2+x+6+x^2-1}{2\left(x+5\right)}=\dfrac{x+5}{2\left(x+5\right)}=\dfrac{1}{2}\)

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

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

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

\(=\dfrac{6x^2-2+x^4-4x^2+3}{\left(x^2-1\right)\left(3x^2-1\right)}:\dfrac{\left(x^2-1\right)\left(3x^2-3\right)-2x^2\left(x^2-3\right)}{x\left(x^2-1\right)\left(3x^2-1\right)}\)

\(=\dfrac{x^4+2x^2+1}{\left(x^2-1\right)\left(3x^2-1\right)}\cdot\dfrac{x\left(x^2-1\right)\left(3x^2-1\right)}{3x^4-6x^2+3-2x^4+6x^2}\)

\(=\dfrac{x\left(x^2+1\right)^2}{x^4+3}\)

a: \(=6x^4-9x^3+3x^2-4x^3+6x^2-2x+10x^2-15x+5\)

\(=6x^4-13x^3+19x^2-17x+5\)

b: \(=6x^4-\dfrac{9}{4}x^3-\dfrac{9}{2}x^2-\dfrac{8}{3}x^3+x^2+2x-\dfrac{20}{3}x^2+\dfrac{5}{2}x+5\)

\(=6x^4-\dfrac{59}{12}x^3-\dfrac{67}{6}x^2+\dfrac{9}{2}x+5\)

c: \(=3x^4-\dfrac{9}{8}x^3-\dfrac{3}{4}x^2+8x^3-3x^2-6x-\dfrac{4}{3}x^2+\dfrac{1}{2}x+1\)

\(=3x^4-\dfrac{55}{8}x^3-\dfrac{25}{12}x^2-\dfrac{11}{2}x+1\)